CHAPTER
3
Kinematics in Two Dimensions
image
A fireboat on the East River in Lower Manhattan, New York, sprays water in a test of its equipment. The arching streams of water follow parabolic paths whose sizes depend on the launch velocity of the water and the acceleration due to gravity, assuming that the effects of air resistance are negligible.�(� Siegfried Layda/Getty Images, Inc.)

3.1� Displacement, Velocity, and Acceleration
In Chapter 2 the concepts of displacement, velocity, and acceleration are used to describe an object moving in one dimension. There are also situations in which the motion is along a curved path that lies in a plane. Such two-dimensional motion can be described using the same concepts. In Grand Prix racing, for example, the course follows a curved road, and Figure 3.1 shows a race car at two different positions along it. These positions are identified by the vectors and , which are drawn from an arbitrary coordinate origin. The displacement of the car is the vector drawn from the initial position at time to the final position at time t. The magnitude of is the shortest distance between the two positions. In the drawing, the vectors and are drawn tail to head, so it is evident that is the vector sum of and . (See Sections 1.5 and 1.6 for a review of vectors and vector addition.) This means that , or
The displacement here is defined as it is in Chapter 2. Now, however, the displacement vector may lie anywhere in a plane, rather than just along a straight line.
image
Figure�3.1�
The displacement of the car is a vector that points from the initial position of the car at time to the final position at time t. The magnitude of is the shortest distance between the two positions.
The average velocity of the car in Figure 3.1 between the two positions is defined in a manner similar to that in Equation 2.2, as the displacement divided by the elapsed time :
��(3.1)
Since both sides of Equation 3.1 must agree in direction, the average velocity vector has the same direction as the displacement . The velocity of the car at an instant of time is its instantaneous velocity . The average velocity becomes equal to the instantaneous velocity in the limit that becomes infinitesimally small :
Figure 3.2 illustrates that the instantaneous velocity is tangent to the path of the car. The drawing also shows the vector components and of the velocity, which are parallel to the x and y axes, respectively.
image
Figure�3.2�
The instantaneous velocity and its two vector components and .
The average acceleration is defined just as it is for one-dimensional motion—namely, as the change in velocity, , divided by the elapsed time :
��(3.2)
The average acceleration has the same direction as the change in velocity . In the limit that the elapsed time becomes infinitesimally small, the average acceleration becomes equal to the instantaneous acceleration :
The acceleration has a vector component along the x direction and a vector component along the y direction.
Check Your Understanding�
1.��
Suppose you are driving due east, traveling a distance of 1500 m in 2 minutes. You then turn due north and travel the same distance in the same time. What can be said about the average speeds and the average velocities for the two segments of the trip?
  1. The average speeds are the same, and the average velocities are the same.
  2. The average speeds are the same, but the average velocities are different.
  3. The average speeds are different, but the average velocities are the same.
Answer:
b.
The average speeds are the same, but the average velocities are different.
3.2� Equations of Kinematics in Two Dimensions
To understand how displacement, velocity, and acceleration are applied to two-dimensional motion, consider a spacecraft equipped with two engines that are mounted perpendicular to each other. These engines produce the only forces that the craft experiences, and the spacecraft is assumed to be at the coordinate origin when , so that . At a later time t, the spacecraft's displacement is . Relative to the x and y axes, the displacement has vector components of and , respectively.
In Figure 3.3 only the engine oriented along the x direction is firing, and the vehicle accelerates along this direction. It is assumed that the velocity in the y direction is zero, and it remains zero, since the y engine is turned off. The motion of the spacecraft along the x direction is described by the five kinematic variables x, , , , and t. Here the symbol “x” reminds us that we are dealing with the x components of the displacement, velocity, and acceleration vectors. (See Sections 1.7 and 1.8 for a review of vector components.) The variables x, , , and are scalar components (or “components,” for short). As Section 1.7 discusses, these components are positive or negative numbers (with units), depending on whether the associated vector components point in the or the direction. If the spacecraft has a constant acceleration along the x direction, the motion is exactly like that described in Chapter 2, and the equations of kinematics can be used. For convenience, these equations are written in the left column of Table 3.1.
image
Figure�3.3�
The spacecraft is moving with a constant acceleration parallel to the x axis. There is no motion in the y direction, and the y engine is turned off.
Table�3.1���
Equations of Kinematics for Constant Acceleration in Two-Dimensional Motion
x Component
Variable
y Component
x
Displacement
y
Acceleration
Final velocity
Initial velocity
t
Elapsed time
t
��(3.3a)
��(3.3b)
��(3.4a)
��(3.4b)
��(3.5a)
��(3.5b)
��(3.6a)
��(3.6b)
Figure 3.4 is analogous to Figure 3.3, except that now only the y engine is firing, and the spacecraft accelerates along the y direction. Such a motion can be described in terms of the kinematic variables y, , , , and t. And if the acceleration along the y direction is constant, these variables are related by the equations of kinematics, as written in the right column of Table 3.1. Like their counterparts in the x direction, the scalar components, y, , , and , may be positive or negative numbers (with units).
image
Figure�3.4�
The spacecraft is moving with a constant acceleration parallel to the y axis. There is no motion in the x direction, and the x engine is turned off.
Problem-Solving Insight.�
It is important to realize that the x part of the motion occurs exactly as it would if the y part did not occur at all. Similarly, the y part of the motion occurs exactly as it would if the x part of the motion did not exist.
If both engines of the spacecraft are firing at the same time, the resulting motion takes place in part along the x axis and in part along the y axis, as Figure 3.5 illustrates. The thrust of each engine gives the vehicle a corresponding acceleration component. The x engine accelerates the craft in the x direction and causes a change in the x component of the velocity. Likewise, the y engine causes a change in the y component of the velocity. In other words, the x and y motions are independent of each other.
image
Figure�3.5�
The two-dimensional motion of the spacecraft can be viewed as the combination of the separate x and y motions.
The independence of the x and y motions lies at the heart of two-dimensional kinematics. It allows us to treat two-dimensional motion as two distinct one-dimensional motions, one for the x direction and one for the y direction. Everything that we have learned in Chapter 2 about kinematics in one dimension will now be applied separately to each of the two directions. In so doing, we will be able to describe the x and y variables separately and then bring these descriptions together to understand the two-dimensional picture. Examples 1 and 2 take this approach in dealing with a moving spacecraft.
Example� 1� The Displacement of a Spacecraft
In Figure 3.5, the directions to the right and upward are the positive directions. In the x direction, the spacecraft has an initial velocity component of and an acceleration component of . In the y direction, the analogous quantities are and . At a time of , find the x and y components of the spacecraft's displacement.
Reasoning
The motion in the x direction and the motion in the y direction can be treated separately, each as a one-dimensional motion subject to the equations of kinematics for constant acceleration (see Table 3.1). By following this procedure we will be able to determine x and y, which specify the spacecraft's location after an elapsed time of 7.0 s.
Problem-Solving Insight.�
When the motion is two-dimensional, the time variable t has the same value for both the x and y directions.
Solution
The data for the motion in the x direction are listed in the following table:
x-Direction Data
x
t
?
7.0 s
The x component of the craft's displacement can be found by using Equation 3.5a:
The data for the motion in the y direction are listed in the following table:
y-Direction Data
y
t
?
7.0 s
The y component of the craft's displacement can be found by using Equation 3.5b:
After 7.0 s, the spacecraft is 740 m to the right and 390 m above the origin.
Analyzing Multiple-Concept Problems�
Example� 2� The Velocity of a Spacecraft
This example also deals with the spacecraft in Figure 3.5. As in Example 1, the x components of the craft's initial velocity and acceleration are and , respectively. The corresponding y components are and . At a time of , find the spacecraft's final velocity (magnitude and direction).
Reasoning
Figure 3.6 shows the final velocity vector, which has components and and a magnitude v. The final velocity is directed at an angle above the axis. The vector and its components form a right triangle, the hypotenuse being the magnitude of the velocity and the components being the other two sides. Thus, we can use the Pythagorean theorem to determine the magnitude v from values for the components and . We can also use trigonometry to determine the directional angle .
image
Figure�3.6�
The velocity vector has components and and a magnitude v. The magnitude gives the speed of the spacecraft, and the angle gives the direction of travel relative to the positive x direction.
Knowns and Unknowns The data for this problem are listed in the table that follows:
Description
Symbol
Value
Comment
x component of acceleration
x component of initial velocity
y component of acceleration
y component of initial velocity
Time
t
7.0 s
Same time for x and y directions
Unknown Variables
Magnitude of final velocity
v
?
Direction of final velocity
?
Modeling the Problem
icon Final Velocity In Figure 3.6 the final velocity vector and its components and form a right triangle. Applying the Pythagorean theorem to this right triangle shows that the magnitude v of the final velocity is given in terms of the components by Equation 1a at the right. From the right triangle in Figure 3.6 it also follows that the directional angle is given by Equation 1b at the right.
��(1a)
��(1b)
icon The Components of the Final Velocity Values are given for the kinematic variables , , and t in the x direction and for the corresponding variables in the y direction (see the table of knowns and unknowns). For each direction, then, these values allow us to calculate the final velocity components and by using Equation 3.3a and 3.3b from the equations of kinematics.
��(3.3a)
��(3.3b)
These expressions can be substituted into Equations 1a and 1b for the magnitude and direction of the final velocity, as shown at the right.
image
Solution
Algebraically combining the results of each step, we find that
image
With the data given for the kinematic variables in the x and y directions, we find that the magnitude and direction of the final velocity of the spacecraft are
After 7.0 s, the spacecraft, at the position determined in Example 1, has a velocity of in a direction of above the positive x axis.
Related Homework: Problem 66
The following Reasoning Strategy gives an overview of how the equations of kinematics are applied to describe motion in two dimensions, as in Examples 1 and 2.
Reasoning Strategy� Applying the Equations of Kinematics in Two Dimensions
1.��
Make a drawing to represent the situation being studied.
2.��
Decide which directions are to be called positive and negative relative to a conveniently chosen coordinate origin. Do not change your decision during the course of a calculation.
3.��
Remember that the time variable t has the same value for the part of the motion along the x axis and the part along the y axis.
4.��
In an organized way, write down the values (with appropriate and signs) that are given for any of the five kinematic variables associated with the x direction and the y direction. Be on the alert for implied data, such as the phrase “starts from rest,” which means that the values of the initial velocity components are zero: and . The data summary tables that are used in the examples are a good way of keeping track of this information. In addition, identify the variables that you are being asked to determine.
5.��
Before attempting to solve a problem, verify that the given information contains values for at least three of the kinematic variables. Do this for the x and the y direction of the motion. Once the three known variables are identified along with the desired unknown variable, the appropriate relations from Table 3.1 can be selected.
6.��
When the motion is divided into segments, remember that the final velocity for one segment is the initial velocity for the next segment.
7.��
Keep in mind that a kinematics problem may have two possible answers. Try to visualize the different physical situations to which the answers correspond.
Check Your Understanding�
2.��
A power boat, starting from rest, maintains a constant acceleration. After a certain time t, its displacement and velocity are and . At time , what would be its displacement and velocity, assuming the acceleration remains the same?
  1. and
  2. and
  3. and
  4. and
Answer:
c.
and
3.3� Projectile Motion
The biggest thrill in baseball is a home run. The motion of the ball on its curving path into the stands is a common type of two-dimensional motion called “projectile motion.” A good description of such motion can often be obtained with the assumption that air resistance is absent.
Using the equations in Table 3.1, we consider the horizontal and vertical parts of the motion separately. In the horizontal or x direction, the moving object (the projectile) does not slow down in the absence of air resistance. Thus, the x component of the velocity remains constant at its initial value or , and the x component of the acceleration is . In the vertical or y direction, however, the projectile experiences the effect of gravity. As a result, the y component of the velocity is not constant and changes. The y component of the acceleration is the downward acceleration due to gravity. If the path or trajectory of the projectile is near the earth's surface, has a magnitude of . In this text, then, the phrase “projectile motion” means that and equals the acceleration due to gravity. Example 3 and other examples in this section illustrate how the equations of kinematics are applied to projectile motion.
Example� 3� A Falling Care Package
Figure 3.7 shows an airplane moving horizontally with a constant velocity of at an altitude of 1050 m. The directions to the right and upward have been chosen as the positive directions. The plane releases a “care package” that falls to the ground along a curved trajectory. Ignoring air resistance, determine the time required for the package to hit the ground.
image
Figure�3.7�
The package falling from the plane is an example of projectile motion, as Examples 3 and 4 discuss.
Reasoning
The time required for the package to hit the ground is the time it takes for the package to fall through a vertical distance of 1050 m. In falling, it moves to the right, as well as downward, but these two parts of the motion occur independently. Therefore, we can focus solely on the vertical part. We note that the package is moving initially in the horizontal or x direction, not in the y direction, so that . Furthermore, when the package hits the ground, the y component of its displacement is , as the drawing shows. The acceleration is that due to gravity, so . These data are summarized as follows:
y-Direction Data
y
t
?
With these data, Equation 3.5b can be used to find the fall time.
Problem-Solving Insight.�
The variables y, , , and are scalar components. Therefore, an algebraic sign (+ or ) must be included with each one to denote direction.
Solution
Since , it follows from Equation 3.5b that and
The freely falling package in Example 3 picks up vertical speed on the way downward. The horizontal component of the velocity, however, retains its initial value of throughout the entire descent. Since the plane also travels at a constant horizontal velocity of , it remains directly above the falling package. The pilot always sees the package directly beneath the plane, as the dashed vertical lines in Figure 3.7 show. This result is a direct consequence of the fact that the package has no acceleration in the horizontal direction. In reality, air resistance would slow down the package, and it would not remain directly beneath the plane during the descent.
Figure 3.8 illustrates what happens to two packages that are released simultaneously from the same height, in order to emphasize that the vertical and horizontal parts of the motion in Example 3 occur independently. Package A is dropped from a stationary balloon and falls straight downward toward the ground, since it has no horizontal velocity component . Package B, on the other hand, is given an initial velocity component of in the horizontal direction, as in Example 3, and follows the path shown in the figure. Both packages hit the ground at the same time. Not only do the packages in Figure 3.8 reach the ground at the same time, but the y components of their velocities are also equal at all points on the way down. However, package B does hit the ground with a greater speed than does package A. Remember, speed is the magnitude of the velocity vector, and the velocity of B has an x component, whereas the velocity of A does not. The magnitude and direction of the velocity vector for package B at the instant just before the package hits the ground is computed in Example 4.
image
Figure�3.8�
Package A and package B are released simultaneously at the same height and strike the ground at the same time because their y variables are the same.
Analyzing Multiple-Concept Problems�
Example� 4� The Velocity of the Care Package
Figure 3.7 shows a care package falling from a plane, and Figure 3.8 shows this package as package B. As in Example 3, the directions to the right and upward are chosen as the positive directions, and the plane is moving horizontally with a constant velocity of at an altitude of 1050 m. Ignoring air resistance, find the magnitude v and the directional angle of the final velocity vector that the package has just before it strikes the ground.
Reasoning
Figures 3.7 and 3.8 show the final velocity vector, which has components and and a magnitude v. The vector is directed at an angle below the horizontal or x direction. We note the right triangle formed by the vector and its components. The hypotenuse of the triangle is the magnitude of the velocity, and the components are the other two sides. As in Example 2, we can use the Pythagorean theorem to express the magnitude or speed v in terms of the components and , and we can use trigonometry to determine the directional angle .
Knowns and Unknowns The data for this problem are listed in the table that follows:
Description
Symbol
Value
Comment
Explicit Data, x Direction
x component of initial velocity
Package has plane's horizontal velocity at instant of release
Implicit Data, x Direction
x component of acceleration
No horizontal acceleration, since air resistance is ignored
Explicit Data, y Direction
y component of displacement
y
Negative, since upward is positive and package falls downward
Implicit Data, y Direction
y component of initial velocity
Package traveling horizontally in x direction at instant of release, not in y direction
y component of acceleration
Acceleration vector for gravity points downward in the negative direction
Unknown Variables
Magnitude of final velocity
v
?
Direction of final velocity
?
Problem-Solving Insight.�
The speed of a projectile at any location along its path is the magnitude v of its velocity at that location: . Both the horizontal and vertical velocity components contribute to the speed.
Modeling the Problem
icon Final Velocity Using the Pythagorean theorem to express the speed v in terms of the components and (see Figure 3.7 or 3.8), we obtain Equation 1a at the right. Furthermore, in a right triangle, the cosine of an angle is the side adjacent to the angle divided by the hypotenuse. With this in mind, we see in Figure 3.7 or Figure 3.8 that the directional angle is given by Equation 1b at the right.
��(1a)
��(1b)
icon The Components of the Final Velocity Reference to the table of knowns andunknowns shows that, in the x direction, values are available for the kinematic variables and . Since the acceleration is zero, the final velocity component remains unchanged from its initial value of , so we have
In the y direction, values are available for y, , and , so that we can determine the final velocity component using Equation 3.6b from the equations of kinematics:
��(3.6b)
These results for and can be substituted into Equations 1a and 1b, as shown at the right.
image
Solution
Algebraically combining the results of each step, we find that
image
With the data given for the kinematic variables in the x and y directions, we find that the magnitude and direction of the final velocity of the package are
Related Homework: Problems 33, 38, 71
An important feature of projectile motion is that there is no acceleration in the horizontal, or x, direction. Conceptual Example 5 discusses an interesting implication of this feature.
Conceptual Example� 5� I Shot a Bullet Into the Air …
Suppose you are driving in a convertible with the top down. The car is moving to the right at a constant velocity. As Figure 3.9 illustrates, you point a rifle straight upward and fire it. In the absence of air resistance, would the bullet land
(a)��
behind you,
(b)��
ahead of you, or
(c)��
in the barrel of the rifle?
image
Figure�3.9�
The car is moving with a constant velocity to the right, and the rifle is pointed straight up. In the absence of air resistance, a bullet fired from the rifle has no acceleration in the horizontal direction. Example 5 discusses what happens to the bullet.
Reasoning
Because there is no air resistance to slow it down, the bullet experiences no horizontal acceleration. Thus, the bullet's horizontal velocity component does not change, and it stays the same as that of the rifle and the car.
Answers (a) and (b) are incorrect. If air resistance were present, it would slow down the bullet and cause it to land behind you, toward the rear of the car. However, air resistance is absent. If the bullet were to land ahead of you, its horizontal velocity component would have to be greater than that of the rifle and the car. This cannot be, since the bullet's horizontal velocity component never changes.
Answer (c) is correct. Since the bullet's horizontal velocity component does not change, it retains its initial value, and remains matched to that of the rifle and the car. As a result, the bullet remains directly above the rifle at all times and would fall directly back into the barrel of the rifle. This situation is analogous to that in Figure 3.7, where the care package, as it falls, remains directly below the plane.
Related Homework: Problem 39
Often projectiles, such as footballs and baseballs, are sent into the air at an angle with respect to the ground. From a knowledge of the projectile's initial velocity, a wealth of information can be obtained about the motion. For instance, Example 6 demonstrates how to calculate the maximum height reached by the projectile.
Example� 6� The Height of a Kickoff
A placekicker kicks a football at an angle of above the horizontal axis, as Figure 3.10 shows. The initial speed of the ball is . Ignore air resistance, and find the maximum height H that the ball attains.
image
Figure�3.10�
A football is kicked with an initial speed of at an angle of θ above the ground. The ball attains a maximum height H and a range R.
Reasoning
The maximum height is a characteristic of the vertical part of the motion, which can be treated separately from the horizontal part. In preparation for making use of this fact, we calculate the vertical component of the initial velocity:
The vertical component of the velocity, , decreases as the ball moves upward. Eventually, at the maximum height H. The data below can be used in Equation 3.6b to find the maximum height:
y-Direction Data
y
t
Problem-Solving Insight.�
When a projectile reaches maximum height, the vertical component of its velocity is momentarily zero . However, the horizontal component of its velocity is not zero.
Solution
From Equation 3.6b, we find that
The height H depends only on the y variables; the same height would have been reached had the ball been thrown straight up with an initial velocity of .
It is also possible to find the total time or “hang time” during which the football in Figure 3.10 is in the air. Example 7 shows how to determine this time.
Example� 7� The Physics of the “Hang Time” of a Football
For the motion illustrated in Figure 3.10, ignore air resistance and use the data from Example 6 to determine the time of flight between kickoff and landing.
Reasoning
Given the initial velocity, it is the acceleration due to gravity that determines how long the ball stays in the air. Thus, to find the time of flight we deal with the vertical part of the motion. Since the ball starts at and returns to ground level, the displacement in the y direction is zero. The initial velocity component in the y direction is the same as that in Example 6; that is, . Therefore, we have
y-Direction Data
y
t
0 m
?
The time of flight can be determined from Equation 3.5b .
Solution
Using Equation 3.5b and the fact that , we find that
There are two solutions to this equation. One is given by , with the result that
The other is given by . The solution we seek is , because corresponds to the initial kickoff.
Another important feature of projectile motion is called the “range.” The range, as Figure 3.10 shows, is the horizontal distance traveled between launching and landing, assuming the projectile returns to the same vertical level at which it was fired. Example 8 shows how to obtain the range.
Example� 8� The Range of a Kickoff
For the motion shown in Figure 3.10 and discussed in Examples 6 and 7, ignore air resistance and calculate the range R of the projectile.
Reasoning
The range is a characteristic of the horizontal part of the motion. Thus, our starting point is to determine the horizontal component of the initial velocity:
Recall from Example 7 that the time of flight is . Since there is no acceleration in the x direction, remains constant, and the range is simply the product of and the time.
Solution
The range is
The range in the previous example depends on the angle at which the projectile is fired above the horizontal. When air resistance is absent, the maximum range results when .
MATH SKILLS�
To show that a projectile launched from and returning to ground level has its maximum range when , we begin with the expression for the range R from Example 8:
Example 7 shows that the time of flight of the projectile is
According to Example 6, the velocity component in this result is , so that
Substituting this expression for t into the range expression gives
Equation (Other Trigonometric Identities) in Appendix E.2 shows that , so the range expression becomes
In this result R has its maximum value when has its maximum value of 1. This occurs when , or when .
The examples considered thus far have used information about the initial location and velocity of a projectile to determine the final location and velocity. Example 9 deals with the opposite situation and illustrates how the final parameters can be used with the equations of kinematics to determine the initial parameters.
Analyzing Multiple-Concept Problems�
Example� 9� A Home Run
A baseball player hits a home run, and the ball lands in the left-field seats, 7.5 m above the point at which it was hit. It lands with a velocity of at an angle of below the horizontal (see Figure 3.11). The positive directions are upward and to the right in the drawing. Ignoring air resistance, find the magnitude and direction of the initial velocity with which the ball leaves the bat.
image
Figure�3.11�
The velocity and location of the baseball upon landing can be used to determine its initial velocity, as Example 9 illustrates.
Reasoning
Just after the ball is hit, its initial velocity has a magnitude and components of and and is directed at an angle above the horizontal or x direction. Figure 3.11 shows the initial velocity vector and its components. As usual, we will use the Pythagorean theorem to relate to and and will use trigonometry to determine .
Knowns and Unknowns The data for this problem are listed in the table that follows:
Description
Symbol
Value
Comment
Explicit Data
y component of displacement
y
Positive, since upward is positive and ball lands above its starting point
Magnitude of final velocity
v
Direction of final velocity
Below the horizontal (see Figure 3.11)
Implicit Data
x component of acceleration
No horizontal acceleration, since air resistance is ignored
y component of acceleration
Acceleration vector for gravity points downward in the negative direction
Unknown Variables
Magnitude of initial velocity
?
Direction of initial velocity
?
Modeling the Problem
icon Initial Velocity The magnitude of the initial velocity can be related to its components and by using the Pythagorean theorem, since the components are perpendicular to one another. This leads to Equation 1a at the right. Referring to Figure 3.11, we can also use trigonometry to express the directional angle in terms of the components and . Thus, we obtain Equation 1b at the right.
��(1a)
��(1b)
icon Component of the Initial Velocity To obtain , we note that the acceleration is zero, since air resistance is being ignored. With no acceleration in the x direction, remains unchanged throughout the motion of the ball. Thus, must equal , the x component of the ball's final velocity. We have, then, that
This result can be substituted into Equations 1a and 1b, as shown at the right.
image
icon Component of the Initial Velocity In contrast to the argument in Step 2, does not equal , the y component of the ball's final velocity, since the ball accelerates in the vertical direction. However, we can use Equation 3.6b from the equations of kinematics to determine :
��(3.6b)
The plus sign is chosen for the square root, since the ball's initial velocity component points upward in Figure 3.11. In Equation 3.6b, can be written as
where the minus sign is present because points downward in the direction in Figure 3.11. We find, then, that
This result can also be substituted into Equations 1a and 1b, as shown at the right.
image
MATH SKILLS�
Tan is the tangent function and is defined as (Equation 1.3), where is the length of the side of a right triangle opposite the angle , and is the length of the side adjacent to the angle (see part a of the drawing). The first step in applying such a trigonometric function is to identify this angle and its associated right triangle. The left side of Figure 3.11 establishes the angle and is reproduced in Figure 3.12b. The shaded triangles in Figure 3.12 reveal that and . Thus, , so that is the angle whose tangent is . This result can be expressed by using the inverse tangent function :
image
Figure�3.12�
Math Skills drawing.
Solution
Algebraically combining the results of each step, we find that
image
With the data given in the table of knowns and unknowns, we find that the magnitude and direction of the ball's initial velocity are
Related Homework: Problems 32, 40
In projectile motion, the magnitude of the acceleration due to gravity affects the trajectory in a significant way. For example, a baseball or a golf ball would travel much farther and higher on the moon than on the earth, when launched with the same initial velocity. The reason is that the moon's gravity is only about one-sixth as strong as the earth's.
Section 2.6 points out that certain types of symmetry with respect to time and speed are present for freely falling bodies. These symmetries are also found in projectile motion, since projectiles are falling freely in the vertical direction. In particular, the time required for a projectile to reach its maximum height H is equal to the time spent returning to the ground. In addition, Figure 3.13 shows that the speed v of the object at any height above the ground on the upward part of the trajectory is equal to the speed v at the same height on the downward part. Although the two speeds are the same, the velocities are different, because they point in different directions. Conceptual Example 10 shows how to use this type of symmetry in your reasoning.
image
Figure�3.13�
The speed v of a projectile at a given height above the ground is the same on the upward and downward parts of the trajectory. The velocities are different, however, since they point in different directions.
Conceptual Example� 10� Two Ways to Throw a Stone
From the top of a cliff overlooking a lake, a person throws two stones. The stones have identical initial speeds , but stone 1 is thrown downward at an angle below the horizontal, while stone 2 is thrown upward at the same angle above the horizontal, as Figure 3.14 shows. Neglect air resistance and decide which stone, if either, strikes the water with the greater velocity: (a) both stones strike the water with the same velocity, (b) stone 1 strikes with the greater velocity, (c) stone 2 strikes with the greater velocity.
image
Figure�3.14�
Two stones are thrown off the cliff with identical initial speeds , but at equal angles θ that are below and above the horizontal. Conceptual Example 10 compares the velocities with which the stones hit the water below.
Reasoning
Note point P in the drawing, where stone 2 returns to its initial height; here the speed of stone 2 is , (the same as its initial speed), but its velocity is directed at an angle below the horizontal. This is exactly the type of projectile symmetry illustrated in Figure 3.13, and this symmetry will lead us to the correct answer.
Answers (b) and (c) are incorrect. You might guess that stone 1, being hurled downward, would strike the water with the greater velocity. Or, you might think that stone 2, having reached a greater height than stone 1, would hit the water with the greater velocity. To understand why neither of these answers is correct, see the response for answer (a) below.
Answer (a) is correct. Let's follow the path of stone 2 as it rises to its maximum height and falls back to earth. When it reaches point P in the drawing, stone 2 has a velocity that is identical to the velocity with which stone 1 is thrown downward from the top of the cliff (see the drawing). From this point on, the velocity of stone 2 changes in exactly the same way as that for stone 1, so both stones strike the water with the same velocity.
Related Homework: Problems 23, 75
In all the examples in this section, the projectiles follow a curved trajectory. In general, if the only acceleration is that due to gravity, the shape of the path can be shown to be a parabola.
Check Your Understanding�
3.��
A projectile is fired into the air, and it follows the parabolic path shown in the drawing, landing on the right. There is no air resistance. At any instant, the projectile has a velocity and an acceleration . Which one or more of the drawings could not represent the directions for and at any point on the trajectory?
image
Answer:
a and c
4.��
An object is thrown upward at an angle above the ground, eventually returning to earth.
(a)��
Is there any place along the trajectory where the velocity and acceleration are perpendicular? If so, where?
Answer:
Yes; when the object is at its highest point.
(b)��
Is there any place where the velocity and acceleration are parallel? If so, where?
Answer:
No.
5.��
Is the acceleration of a projectile equal to zero when the projectile reaches the top of its trajectory?
Answer:
No.
6.��
In baseball, the pitcher's mound is raised to compensate for the fact that the ball falls downward as it travels from the pitcher toward the batter. If baseball were played on the moon, would the pitcher's mound have to be
  1. higher than,
  2. lower than, or
  3. the same height as it is on earth?
Answer:
b.
lower than, or
7.��
A tennis ball is hit upward into the air and moves along an arc. Neglecting air resistance, where along the arc is the speed of the ball
(a)��
a minimum and
Answer:
when the ball is at its highest point in the trajectory
(b)��
a maximum?
Answer:
at the initial and final positions of the motion
8.��
A wrench is accidentally dropped from the top of the mast on a sailboat. Air resistance is negligible. Will the wrench hit at the same place on the deck whether the sailboat is at rest or moving with a constant velocity?
Answer:
Yes.
9.��
A rifle, at a height H above the ground, fires a bullet parallel to the ground. At the same instant and at the same height, a second bullet is dropped from rest. In the absence of air resistance, which bullet, if either, strikes the ground first?
Answer:
Both bullets reach the ground at the same time.
10.��
A stone is thrown horizontally from the top of a cliff and eventually hits the ground below. A second stone is dropped from rest from the same cliff, falls through the same height, and also hits the ground below. Ignore air resistance. Is each of the following quantities different or the same in the two cases?
(a)��
Displacement
Answer:
The displacement is greater for the stone thrown horizontally.
(b)��
Speed just before impact with the ground
Answer:
The impact speed is greater for the stone thrown horizontally.
(c)��
Time of flight
Answer:
The time of flight is the same for both stones.
11.��
A leopard springs upward at a angle and then falls back to the ground. Air resistance is negligible. Does the leopard, at any point on its trajectory, ever have a speed that is one-half its initial value?
Answer:
No.
12.��
Two balls are launched upward from the same spot at different angles with respect to the ground. Both balls rise to the same maximum height. Ball A, however, follows a trajectory that has a greater range than that of ball B. Ignoring air resistance, decide which ball, if either, has the greater launch speed.
Answer:
Ball A has the greater launch speed.
3.4�
* This equation assumes that the train and the ground move in a straight line relative to one another.
Relative Velocity
To someone hitchhiking along a highway, two cars speeding by in adjacent lanes seem like a blur. But if the cars have the same velocity, each driver sees the other remaining in place, one lane away. The hitchhiker observes a velocity of perhaps , but each driver observes the other's velocity to be zero. Clearly, the velocity of an object is relative to the observer who is making the measurement.
image
In landing on a moving aircraft carrier, the pilot of the helicopter must match the helicopter's horizontal velocity to the carrier's velocity, so that the relative velocity of the helicopter and the carrier is zero.�(� Tal Cohen/AFP/Getty Images, Inc.)
Figure 3.15 illustrates the concept of relative velocity by showing a passenger walking toward the front of a moving train. The people sitting on the train see the passenger walking with a velocity of , where the plus sign denotes a direction to the right. Suppose the train is moving with a velocity of relative to an observer standing on the ground. Then the ground-based observer would see the passenger moving with a velocity of , due in part to the walking motion and in part to the train's motion. As an aid in describing relative velocity, let us define the following symbols:
In terms of these symbols, the situation in Figure 3.15 can be summarized as follows:
��(3.7)
or
According to Equation 3.7*, is the vector sum of and , and this sum is shown in the drawing. Had the passenger been walking toward the rear of the train, rather than toward the front, the velocity relative to the ground-based observer would have been .
image
Figure�3.15�
The velocity of the passenger relative to the ground-based observer is .It is the vector sum of the velocity of the passenger relative to the train and the velocity of the train relative to the ground:.
Each velocity symbol in Equation 3.7 contains a two-letter subscript. The first letter in the subscript refers to the body that is moving, while the second letter indicates the object relative to which the velocity is measured. For example, and are the velocities of the Train and Passenger measured relative to the Ground. Similarly, is the velocity of the Passenger measured by an observer sitting on the Train.
The ordering of the subscript symbols in Equation 3.7 follows a definite pattern. The first subscript (P) on the left side of the equation is also the first subscript on the right side of the equation. Likewise, the last subscript (G) on the left side is also the last subscript on the right side. The third subscript (T) appears only on the right side of the equation as the two “inner” subscripts. The colored boxes below emphasize the pattern of the symbols in the subscripts:
In other situations, the subscripts will not necessarily be P, G, and T, but will be compatible with the names of the objects involved in the motion.
Equation 3.7 has been presented in connection with one-dimensional motion, but the result is also valid for two-dimensional motion. Figure 3.16 depicts a common situation that deals with relative velocity in two dimensions. Part a of the drawing shows a boat being carried downstream by a river; the engine of the boat is turned off. In part b, the engine is turned on, and now the boat moves across the river in a diagonal fashion because of the combined motion produced by the current and the engine. The list below gives the velocities for this type of motion and the objects relative to which they are measured:
The velocity of the boat relative to the water is the velocity measured by an observer who, for instance, is floating on an inner tube and drifting downstream with the current. When the engine is turned off, the boat also drifts downstream with the current, and is zero. When the engine is turned on, however, the boat can move relative to the water, and is no longer zero. The velocity of the water relative to the shore is the velocity of the current measured by an observer on the shore. The velocity of the boat relative to the shore is due to the combined motion of the boat relative to the water and the motion of the water relative to the shore. In symbols,
The ordering of the subscripts in this equation is identical to that in Equation 3.7, although the letters have been changed to reflect a different physical situation. Example 11 illustrates the concept of relative velocity in two dimensions.
image
Figure�3.16�
(a) A boat with its engine turned off is carried along by the current. (b) With the engine turned on, the boat moves across the river in a diagonal fashion.
Example� 11� Crossing a River
The engine of a boat drives it across a river that is 1800 m wide. The velocity of the boat relative to the water is , directed perpendicular to the current, as in Figure 3.17. The velocity of the water relative to the shore is . (a) What is the velocity of the boat relative to the shore? (b) How long does it take for the boat to cross the river?
image
Figure�3.17�
The velocity of the boat relative to the shore is . It is the vector sum of the velocity of the boat relative to the water and the velocity of the water relative to the shore:.
Reasoning
(a) The velocity of the boat relative to the shore is . It is the vector sum of the velocity of the boat relative to the water and the velocity of the water relative to the shore:. Since and are both known, we can use this relation among the velocities, with the aid of trigonometry, to find the magnitude and directional angle of .(b) The component of that is parallel to the width of the river (see Figure 3.17) determines how fast the boat crosses the river; this parallel component is . The time for the boat to cross the river is equal to the width of the river divided by the magnitude of this velocity component.
Solution
(a)��
Since the vectors and are perpendicular (see Figure 3.17), the magnitude of can be determined by using the Pythagorean theorem:
Thus, the boat moves at a speed of with respect to an observer on shore. The direction of the boat relative to the shore is given by the angle in the drawing:
(b)��
The time t for the boat to cross the river is
Sometimes, situations arise when two vehicles are in relative motion, and it is useful to know the relative velocity of one with respect to the other. Example 12 considers this type of relative motion.
Example� 12� Approaching an Intersection
Figure 3.18a shows two cars approaching an intersection along perpendicular roads. The cars have the following velocities:
Find the magnitude and direction of , where
image
Figure�3.18�
Two cars are approaching an intersection along perpendicular roads.
Reasoning
To find , we use an equation whose subscripts follow the order outlined earlier. Thus,
In this equation, the term is the velocity of the ground relative to a passenger in car B, rather than , which is given as , northward. In other words, the subscripts are reversed. However, is related to according to
This relationship reflects the fact that a passenger in car B, moving northward relative to the ground, looks out the car window and sees the ground moving southward, in the opposite direction. Therefore, the equation may be used to find , provided we recognize as a vector that points opposite to the given velocity . With this in mind, Figure 3.18b illustrates how and are added vectorially to give .
Problem-Solving Insight.�
In general, the velocity of object R relative to object S is always the negative of the velocity of object S relative to R: .
Solution
From the vector triangle in Figure 3.18b, the magnitude and direction of can be calculated as
and
The physics of raindrops falling on car windows. While driving a car, have you ever noticed that the rear window sometimes remains dry, even though rain is falling? This phenomenon is a consequence of relative velocity, as Figure 3.19 helps to explain. Part a shows a car traveling horizontally with a velocity of and a raindrop falling vertically with a velocity of . Both velocities are measured relative to the ground. To determine whether the raindrop hits the window, however, we need to consider the velocity of the raindrop relative to the car, not to the ground. This velocity is , and we know that
Here, we have used the fact that . Part b of the drawing shows the tail-to-head arrangement corresponding to this vector subtraction and indicates that the direction of is given by the angle . In comparison, the rear window is inclined at an angle with respect to the vertical (see the blowup in part a). When is greater than , the raindrop will miss the window. However, is determined by the speed of the raindrop and the speed of the car, according to . At higher car speeds, the angle becomes too large for the drop to hit the window. At a high enough speed, then, the car simply drives out from under each falling drop!
image
Figure�3.19�
(a) With respect to the ground, a car is traveling at a velocity of and a raindrop is falling at a velocity of . The rear window of the car is inclined at an angle with respect to the vertical. (b) This tail-to-head arrangement of vectors corresponds to the equation .
Check Your Understanding�
13.��
Three cars, A, B, and C, are moving along a straight section of a highway. The velocity of A relative to B is , the velocity of A relative to C is , and the velocity of C relative to B is . Fill in the missing velocities in the table.
(a)
?
(b)
?
(c)
?
(d)
?
Answer:
(a)��
(b)��
(c)��
(d)��
14.��
On a riverboat cruise, a plastic bottle is accidentally dropped overboard. A passenger on the boat estimates that the boat pulls ahead of the bottle by 5 meters each second. Is it possible to conclude that the magnitude of the velocity of the boat with respect to the shore is ?
Answer:
No.
15.��
A plane takes off at St. Louis, flies straight to Denver, and then returns the same way. The plane flies at the same speed with respect to the ground during the entire flight, and there are no head winds or tail winds. Since the earth revolves around its axis once a day, you might expect that the times for the outbound trip and the return trip differ, depending on whether the plane flies against the earth's rotation or with it. Is this true, or are the two times the same?
Answer:
The two times are the same.
16.��
A child is playing on the floor of a recreational vehicle (RV) as it moves along the highway at a constant velocity. He has a toy cannon, which shoots a marble at a fixed angle and speed with respect to the floor. The cannon can be aimed toward the front or the rear of the RV. Is the range toward the front the same as, less than, or greater than the range toward the rear? Answer this question
(a)��
from the child's point of view and
Answer:
The range toward the front is the same as the range toward the rear.
(b)��
from the point of view of an observer standing still on the ground.
Answer:
The range toward the front is greater than the range toward the rear.
17.��
Three swimmers can swim equally fast relative to the water. They have a race to see who can swim across a river in the least time. Swimmer A swims perpendicular to the current and lands on the far shore downstream, because the current has swept him in that direction. Swimmer B swims upstream at an angle to the current, choosing the angle so that he lands on the far shore directly opposite the starting point. Swimmer C swims downstream at an angle to the current in an attempt to take advantage of the current. Who crosses the river in the least time?
Answer:
swimmer A
3.5� Concepts & Calculations
A primary focus of this chapter has been projectile motion. This section presents two additional examples that serve as a review of the basic features of this type of motion. Example 13 deals with the fact that projectile motion consists of a horizontal and a vertical part, the central idea being that these two parts occur independently of one another. Example 14 stresses the fact that the time variable has the same value for both the horizontal and vertical parts of the motion.
Concepts & Calculations Example� 13�
Projectile Motion
In a circus act, Biff the clown is fired from a cannon at an initial velocity directed at an angle above the horizontal, as Figure 3.20 shows. Simultaneously, two other clowns are also launched. Bongo is launched horizontally on roller skates at a speed of . He rolls along the ground while Biff flies through the air. When Biff returns to the ground, he lands side by side with his roller-skating friend, who is gliding by just at the instant of landing. The third clown, Bingo, however, is fired straight upward at a speed of and reaches the same maximum height at the same instant as Biff. Ignore air resistance, and assume that the roller skates are unimpeded by friction. Find the speed and the angle for Biff.
image
Figure�3.20�
Example 13 discusses the projectile motion shown here, in which three circus clowns, Biff, Bongo, and Bingo, are launched simultaneously.
Solution
Based on the Concept Questions and Answers, we can identify the x and y components of Biff's initial velocity as follows:
The initial speed is the magnitude of the initial velocity, and it, along with the directional angle , can be determined from the components:
Concept Questions and Answers
Is Bongo's left-to-right motion the same as or different from the horizontal part of Biff's motion along his trajectory?
Answer:
The horizontal and vertical parts of projectile motion occur independently of one another. Therefore, since Biff lands side by side with Bongo, Bongo's left-to-right motion is identical to the horizontal part of Biff's motion along his trajectory.
Is Biff's motion in the horizontal direction determined by the initial velocity , just its horizontal component , or just its vertical component ?
Answer:
Just the horizontal component determines Biff's motion in the horizontal direction.
Is Bingo's up-and-down motion the same as or different from the vertical part of Biff's motion along his trajectory?
Answer:
The horizontal and vertical parts of projectile motion occur independently of one another. Therefore, since they reach the same maximum height at the same instant, Bingo's up-and-down motion is identical to the vertical part of Biff's motion along his trajectory.
Is Biff's initial motion in the vertical direction determined by the initial velocity , just its horizontal component , or just its vertical component ?
Answer:
Just the vertical component determines Biff's motion in the vertical direction.
Concepts & Calculations Example� 14�
Time and Projectile Motion
A projectile is launched from and returns to ground level, as Figure 3.21 shows. Air resistance is absent. The horizontal range of the projectile is measured to be , and the horizontal component of the launch velocity is . Find the vertical component of the launch velocity.
image
Figure�3.21�
A projectile, launched with a velocity whose horizontal component is , has a range of . From these data the vertical component of the initial velocity can be determined.
Solution
From the constant-velocity horizontal motion, we find that the time is
For the vertical part of the motion, we know that the displacement is zero and that the acceleration due to gravity is , assuming that upward is the positive direction. Therefore, we can use Equation 3.5b to find the initial y component of the velocity:
Concept Questions and Answers
What is the final value of the horizontal component of the projectile's velocity?
Answer:
The final value of the horizontal component of the projectile's velocity is the same as the initial value in the absence of air resistance. In other words, the horizontal motion occurs at a constant velocity of .
Can the time be determined for the horizontal part of the motion?
Answer:
Yes. In constant-velocity motion, the time can be calculated as the horizontal distance (the range) divided by the magnitude of the horizontal component of the projectile's velocity.
Is the time for the horizontal part of the motion the same as the time for the vertical part of the motion?
Answer:
Yes. The value for the time calculated for the horizontal part of the motion can be used to analyze the vertical part of the motion.
For the vertical part of the motion, what is the displacement of the projectile?
Answer:
Since the projectile is launched from and returns to ground level, the vertical displacement is zero.
�� Concept Summary
3.1 Displacement, Velocity, and Acceleration The position of an object is located with a vector drawn from the coordinate origin to the object. The displacement of the object is defined as , where and specify its final and initial positions, respectively.
��(3.1)
The average velocity of an object moving between two positions is defined as its displacement divided by the elapsed time , as in Equation 3.1.
��(1)
The instantaneous velocity is the velocity at an instant of time. The average velocity becomes equal to the instantaneous velocity in the limit that becomes infinitesimally small , as shown in Equation 1.
��(3.2)
The average acceleration of an object is the change in its velocity, divided by the elapsed time , as in Equation 3.2.
��(2)
The instantaneous acceleration is the acceleration at an instant of time. The average acceleration becomes equal to the instantaneous acceleration in the limit that the elapsed time becomes infinitesimally small, as shown in Equation 2.
3.2 Equations of Kinematics in Two Dimensions Motion in two dimensions can be described in terms of the time t and the x and y components of four vectors: the displacement, the acceleration, and the initial and final velocities. The part of the motion occurs exactly as it would if the part did not occur at all. Similarly, the part of the motion occurs exactly as it would if the part of the motion did not exist. The motion can be analyzed by treating the x and y components of the four vectors separately and realizing that the time t is the same for each component.
When the acceleration is constant, the x components of the displacement, the acceleration, and the initial and final velocities are related by the equations of kinematics, and so are the y components:
x Component
y Component
��(3.3a)
��(3.3b)
��(3.4a)
��(3.4b)
��(3.5a)
��(3.5b)
��(3.6a)
��(3.6b)
The directions of the components of the displacement, the acceleration, and the initial and final velocities are conveyed by assigning a plus or minus sign to each one.
3.3 Projectile Motion Projectile motion is an idealized kind of motion that occurs when a moving object (the projectile) experiences only the acceleration due to gravity, which acts vertically downward. If the trajectory of the projectile is near the earth's surface, the vertical component of the acceleration has a magnitude of . The acceleration has no horizontal component , the effects of air resistance being negligible.
There are several symmetries in projectile motion: (1) The time to reach maximum height from any vertical level is equal to the time spent returning from the maximum height to that level. (2) The speed of a projectile depends only on its height above its launch point, and not on whether it is moving upward or downward.
��(3)
3.4 Relative Velocity The velocity of object A relative to object B is , and the velocity of object B relative to object C is . The velocity of A relative to C is shown in Equation 3 (note the ordering of the subscripts). While the velocity of object A relative to object B is , the velocity of B relative to A is .
�� Focus on Concepts
Section 3.3 Projectile Motion
1.��
The drawing shows projectile motion at three points along the trajectory. The speeds at the points are , , and . Assume there is no air resistance and rank the speeds, largest to smallest. (Note that the symbol > means “greater than.”)
  1. image
    Question 1
Answer:
a.
(a) The horizontal component of the projectile's velocity remains constant throughout the motion, since the acceleration in the horizontal direction is zero . The vertical component , however, changes as the projectile moves. This component is greatest at point 1, decreases to zero at point 2 at the top of the trajectory, and then increases to a magnitude less than that at point 1 as the projectile approaches point 3.
3.��
Two balls are thrown from the top of a building, as in the drawing. Ball 1 is thrown straight down, and ball 2 is thrown with the same speed, but upward at an angle with respect to the horizontal. Consider the motion of the balls after they are released. Which one of the following statements is true?
image
  1. The acceleration of ball 1 becomes larger and larger as it falls, because the ball is going faster and faster.
  2. The acceleration of ball 2 decreases as it rises, becomes zero at the top of the trajectory, and then increases as the ball begins to fall toward the ground.
  3. Both balls have the same acceleration at all times.
  4. Ball 2 has an acceleration in both the horizontal and vertical directions, but ball 1 has an acceleration only in the vertical direction.
Answer:
c.
Both balls have the same acceleration at all times.
(c) The acceleration due to gravity is the same for both balls, despite the fact that they have different velocities.
4.��
Each drawing shows three points along the path of a projectile, one on its way up, one at the top, and one on its way down. The launch point is on the left in each drawing. Which drawing correctly represents the acceleration of the projectile at these three points?
image
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
Answer:
d.
4
(d) The acceleration of a projectile is the same at all points on the trajectory. It points downward, toward the earth, and has a magnitude of .
6.��
Ball 1 is thrown into the air and it follows the trajectory for projectile motion shown in the drawing. At the instant that ball 1 is at the top of its trajectory, ball 2 is dropped from rest from the same height. Which ball reaches the ground first?
image
  1. Ball 1 reaches the ground first, since it is moving at the top of the trajectory, while ball 2 is dropped from rest.
  2. Ball 2 reaches the ground first, because it has the shorter distance to travel.
  3. Both balls reach the ground at the same time.
  4. There is not enough information to tell which ball reaches the ground first.
Answer:
d.
There is not enough information to tell which ball reaches the ground first.
(c) The time for a projectile to reach the ground depends only on the component (or vertical component) of its variables, i.e., , , and . These variables are the same for both balls. The fact that Ball 1 is moving horizontally at the top of its trajectory does not play a role in the time it takes for it to reach the ground.
9.��
Two objects are fired into the air, and the drawing shows the projectile motions. Projectile 1 reaches the greater height, but projectile 2 has the greater range. Which one is in the air for the greater amount of time?
image
  1. Projectile 1, because it travels higher than projectile 2.
  2. Projectile 2, because it has the greater range.
  3. Both projectiles spend the same amount of time in the air.
  4. Projectile 2, because it has the smaller initial speed and, therefore, travels more slowly than projectile 1.
Answer:
a.
Projectile 1, because it travels higher than projectile 2.
(a) The time a projectile is in the air is equal to twice the time it takes to fall from its maximum height. Projectile 1 reaches the greater height, so it spends the greater amount of time in the air.
Section 3.4 Relative Velocity
14.��
A slower-moving car is traveling behind a faster-moving bus. The velocities of the two vehicles are as follows:
A passenger on the bus gets up and walks toward the front of the bus with a velocity of , where of the Passenger relative to the . What is , the velocity of the Passenger relative to the Car?
Answer:
c.
(c) The velocity of the passenger relative to the car is given by , according to the subscripting method discussed in Section 3.4. However, the last term on the right of this equation is given by . So, .
15.��
Your car is traveling behind a jeep. Both are moving at the same speed, so the velocity of the jeep relative to you is zero. A spare tire is strapped to the back of the jeep. Suddenly the strap breaks, and the tire falls off the jeep. Will your car hit the spare tire before the tire hits the road? Assume that air resistance is absent.
  1. Yes. As long as the car doesn't slow down, it will hit the tire.
  2. No. The car will not hit the tire before the tire hits the ground, no matter how close you are to the jeep.
  3. If the tire falls from a great enough height, the car will hit the tire.
  4. If the car is far enough behind the jeep, the car will not hit the tire.
Answer:
b.
No. The car will not hit the tire before the tire hits the ground, no matter how close you are to the jeep.
(b) The velocity of the jeep relative to you is zero. Thus, the horizontal component of the tire's velocity relative to you is also zero. Since this component of the velocity never changes as the tire falls, the car cannot hit the tire, regardless of how close the car is to the jeep.
16.��
The drawing shows two cars traveling in different directions with different speeds. Their velocities are:
The passenger of car B looks out the window and sees car A. What is the velocity (magnitude and direction) of car A as observed by the passenger of car B? In other words, what is the velocity of car A relative to car B? Give the directional angle of with respect to due east.
image
Answer:
The magnitude of the velocity of car A relative to car B is . The angle that the velocity makes with respect to due east is south of east.
�� Problems
Section 3.1 Displacement, Velocity, and Acceleration
1.��
�Two trees have perfectly straight trunks and are both growing perpendicular to the flat horizontal ground beneath them. The sides of the trunks that face each other are separated by 1.3 m. A frisky squirrel makes three jumps in rapid succession. First, he leaps from the foot of one tree to a spot that is 1.0 m above the ground on the other tree. Then, he jumps back to the first tree, landing on it at a spot that is 1.7 m above the ground. Finally, he leaps back to the other tree, now landing at a spot that is 2.5 m above the ground. What is the magnitude of the squirrel's displacement?
Answer:
2.8 m
REASONING The displacement is a vector drawn from the initial position to the final position. The magnitude of the displacement is the shortest distance between the positions. Note that it is only the initial and final positions that determine the displacement. The fact that the squirrel jumps to an intermediate position before reaching his final position is not important. The trees are perfectly straight and both growing perpendicular to the flat horizontal ground beneath them. Thus, the distance between the trees and the length of the trunk of the second tree below the squirrel's final landing spot form the two perpendicular sides of a right triangle, as the drawing shows. To this triangle, we can apply the Pythagorean theorem and determine the magnitude of the displacement vector A.
t0001-nu
SOLUTION According to the Pythagorean theorem, we have
2.��
A meteoroid is traveling east through the atmosphere at 18.3 km/s while descending at a rate of 11.5 km/s. What is its speed, in km/s?
3.��
�In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.050 s, during which time it experiences an acceleration of . The ball is launched at an angle of above the ground. Determine the horizontal and vertical components of the launch velocity.
Answer:
4.��
A baseball player hits a triple and ends up on third base. A baseball “diamond” is a square, each side of length 27.4 m, with home plate and the three bases on the four corners. What is the magnitude of the player's displacement?
5.��
�In diving to a depth of 750 m, an elephant seal also moves 460 m due east of his starting point. What is the magnitude of the seal's displacement?
Answer:
REASONING The displacement of the elephant seal has two components; 460 m due east and 750 m downward. These components are mutually perpendicular; hence, the Pythagorean theorem can be used to determine their resultant.
t0002-nu
SOLUTION From the Pythagorean theorem,
Therefore,
6.��
A mountain-climbing expedition establishes two intermediate camps, labeled A and B in the drawing, above the base camp. What is the magnitude of the displacement between camp A and camp B?
image
7.��
�A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be 162 km away. The radar antenna is pointing upward at an angle of from the ground. Find the x and y components (in km) of the position vector of the satellite, relative to the antenna.
Answer:
,
REASONING AND SOLUTION
8.��
�In a mall, a shopper rides up an escalator between floors. At the top of the escalator, the shopper turns right and walks 9.00 m to a store. The magnitude of the shopper's displacement from the bottom of the escalator to the store is 16.0 m. The vertical distance between the floors is 6.00 m. At what angle is the escalator inclined above the horizontal?
9.��
�A skateboarder, starting from rest, rolls down a 12.0-m ramp. When she arrives at the bottom of the ramp her speed is .
(a)��
Determine the magnitude of her acceleration, assumed to be constant.
Answer:
REASONING
(a)��
We designate the direction down and parallel to the ramp as the direction, and the table shows the variables that are known. Since three of the five kinematic variables have values, one of the equations of kinematics can be employed to find the acceleration .
x-Direction Data
?
0 m/s
(b)��
The acceleration vector points down and parallel to the ramp, and the angle of the ramp is relative to the ground (see the drawing). Therefore, trigonometry can be used to determine the component of the acceleration that is parallel to the ground.
t0004-nu
SOLUTION
(a)��
Equation 3.6a can be used to find the acceleration in terms of the three known variables. Solving this equation for gives
(b)��
The drawing shows that the acceleration vector is oriented 25.0 relative to the ground. The component of the acceleration that is parallel to the ground is
(b)��
If the ramp is inclined at with respect to the ground, what is the component of her acceleration that is parallel to the ground?
Answer:
REASONING
(a)��
We designate the direction down and parallel to the ramp as the direction, and the table shows the variables that are known. Since three of the five kinematic variables have values, one of the equations of kinematics can be employed to find the acceleration .
x-Direction Data
?
0 m/s
(b)��
The acceleration vector points down and parallel to the ramp, and the angle of the ramp is relative to the ground (see the drawing). Therefore, trigonometry can be used to determine the component of the acceleration that is parallel to the ground.
t0004-nu
SOLUTION
(a)��
Equation 3.6a can be used to find the acceleration in terms of the three known variables. Solving this equation for gives
(b)��
The drawing shows that the acceleration vector is oriented 25.0 relative to the ground. The component of the acceleration that is parallel to the ground is
REASONING
(a)��
We designate the direction down and parallel to the ramp as the direction, and the table shows the variables that are known. Since three of the five kinematic variables have values, one of the equations of kinematics can be employed to find the acceleration .
x-Direction Data
?
0 m/s
(b)��
The acceleration vector points down and parallel to the ramp, and the angle of the ramp is relative to the ground (see the drawing). Therefore, trigonometry can be used to determine the component of the acceleration that is parallel to the ground.
t0004-nu
SOLUTION
(a)��
Equation 3.6a can be used to find the acceleration in terms of the three known variables. Solving this equation for gives
(b)��
The drawing shows that the acceleration vector is oriented 25.0 relative to the ground. The component of the acceleration that is parallel to the ground is
*10.��
�A bird watcher meanders through the woods, walking 0.50 km due east, 0.75 km due south, and 2.15 km in a direction north of west. The time required for this trip is 2.50 h. Determine the magnitude and direction (relative to due west) of the bird watcher's
(a)��
displacement and
(b)��
average velocity. Use kilometers and hours for distance and time, respectively.
*11.��
�The earth moves around the sun in a nearly circular orbit of radius . During the three summer months (an elapsed time of ), the earth moves one-fourth of the distance around the sun.
(a)��
What is the average speed of the earth?
Answer:
(b)��
What is the magnitude of the average velocity of the earth during this period?
Answer:
Section 3.2 Equations of Kinematics in Two Dimensions
Section 3.3 Projectile Motion
12.��
A spacecraft is traveling with a velocity of along the direction. Two engines are turned on for a time of 842 s. One engine gives the spacecraft an acceleration in the direction of , while the other gives it an acceleration in the direction of . At the end of the firing, find
(a)��
and
(b)��
.
13.��
�A volleyball is spiked so that it has an initial velocity of directed downward at an angle of below the horizontal. What is the horizontal component of the ball's velocity when the opposing player fields the ball?
Answer:
REASONING The vertical component of the ball's velocity changes as the ball approaches the opposing player. It changes due to the acceleration of gravity. However, the horizontal component does not change, assuming that air resistance can be neglected. Hence, the horizontal component of the ball's velocity when the opposing player fields the ball is the same as it was initially.
SOLUTION Using trigonometry, we find that the horizontal component is
14.��
As a tennis ball is struck, it departs from the racket horizontally with a speed of . The ball hits the court at a horizontal distance of 19.6 m from the racket. How far above the court is the tennis ball when it leaves the racket?
15.��
A skateboarder shoots off a ramp with a velocity of , directed at an angle of above the horizontal. The end of the ramp is 1.2 m above the ground. Let the x axis be parallel to the ground, the direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp.
(a)��
How high above the ground is the highest point that the skateboarder reaches?
Answer:
2.8 m
(b)��
When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?
Answer:
2.0 m
16.��
�A puck is moving on an air hockey table. Relative to an x, y coordinate system at time , the x components of the puck's initial velocity and acceleration are and . The y components of the puck's initial velocity and acceleration are and . Find the magnitude and direction of the puck's velocity at a time of . Specify the direction relative to the axis.
17.��
�A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is at an angle of above the table, and it lands on the magazine 0.0770 s after leaving the table. Ignore air resistance. How thick is the magazine? Express your answer in millimeters.
Answer:
9.4 mm
REASONING Since the spider encounters no appreciable air resistance during its leap, it can be treated as a projectile. The thickness of the magazine is equal to the spider's vertical displacement during the leap. The relevant data are as follows (assuming upward to be the direction):
y-Direction Data
?
0.0770 s
SOLUTION We will calculate the spider's vertical displacement directly from (Equation 3.5b):
To express this result in millimeters, we use the fact that 1 m equals 1000 mm:
18.��
A horizontal rifle is fired at a bull's-eye. The muzzle speed of the bullet is . The gun is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.025 m below the center. What is the horizontal distance between the end of the rifle and the bull's-eye?
19.��
�A golfer imparts a speed of to a ball, and it travels the maximum possible distance before landing on the green. The tee and the green are at the same elevation.
(a)��
How much time does the ball spend in the air?
Answer:
4.37 s
(b)��
What is the longest hole in one that the golfer can make, if the ball does not roll when it hits the green?
Answer:
93.5 m
20.��
�A golfer hits a shot to a green that is elevated 3.0 m above the point where the ball is struck. The ball leaves the club at a speed of at an angle of above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
21.��
�In the aerials competition in skiing, the competitors speed down a ramp that slopes sharply upward at the end. The sharp upward slope launches them into the air, where they perform acrobatic maneuvers. The end of a launch ramp is directed above the horizontal. With this launch angle, a skier attains a height of 13 m above the end of the ramp. What is the skier's launch speed?
Answer:
REASONING When the skier leaves the ramp, she exhibits projectile motion. Since we know the maximum height attained by the skier, we can find her launch speed using Equation 3.6b, , where .
SOLUTION At the highest point in her trajectory, . Solving Equation 3.6b for we obtain, taking upward as the positive direction,
22.��
�A space vehicle is coasting at a constant velocity of in the direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at in the direction. After 45.0 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find
(a)��
the magnitude and
(b)��
the direction of the vehicle's velocity relative to the space station. Express the direction as an angle measured from the direction.
23.��
�As preparation for this problem, review Conceptual Example 10. The drawing shows two planes each about to drop an empty fuel tank. At the moment of release each plane has the same speed of , and each tank is at the same height of 2.00 km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of above the horizontal and the other is flying at an angle of below the horizontal. Find the magnitude and direction of the velocity with which the fuel tank hits the ground if it is from
image
(a)��
plane A and
Answer:
, with respect to the horizontal
REASONING Since the magnitude of the velocity of the fuel tank is given by , it is necessary to know the velocity components and just before impact. At the instant of release, the empty fuel tank has the same velocity as that of the plane. Therefore, the magnitudes of the initial velocity components of the fuel tank are given by and , where is the speed of the plane at the instant of release. Since the motion has zero acceleration, the component of the velocity of the plane remains equal to for all later times while the tank is airborne. The component of the velocity of the tank after it has undergone a vertical displacement is given by Equation 3.6b.
SOLUTION
(a)��
Taking up as the positive direction, the velocity components of the fuel tank just before it hits the ground are
From Equation 3.6b, we have
Therefore, the magnitude of the velocity of the fuel tank just before impact is
The velocity vector just before impact is inclined at an angle with respect to the horizontal. This angle is
(b)��
As shown in Conceptual Example 10, once the fuel tank in part a rises and falls to the same altitude at which it was released, its motion is identical to the fuel tank in part b. Therefore, the velocity of the fuel tank in part b just before impact is .
(b)��
plane B. In each part, give the directional angles with respect to the horizontal.
Answer:
, with respect to the horizontal
REASONING Since the magnitude of the velocity of the fuel tank is given by , it is necessary to know the velocity components and just before impact. At the instant of release, the empty fuel tank has the same velocity as that of the plane. Therefore, the magnitudes of the initial velocity components of the fuel tank are given by and , where is the speed of the plane at the instant of release. Since the motion has zero acceleration, the component of the velocity of the plane remains equal to for all later times while the tank is airborne. The component of the velocity of the tank after it has undergone a vertical displacement is given by Equation 3.6b.
SOLUTION
(a)��
Taking up as the positive direction, the velocity components of the fuel tank just before it hits the ground are
From Equation 3.6b, we have
Therefore, the magnitude of the velocity of the fuel tank just before impact is
The velocity vector just before impact is inclined at an angle with respect to the horizontal. This angle is
(b)��
As shown in Conceptual Example 10, once the fuel tank in part a rises and falls to the same altitude at which it was released, its motion is identical to the fuel tank in part b. Therefore, the velocity of the fuel tank in part b just before impact is .
REASONING Since the magnitude of the velocity of the fuel tank is given by , it is necessary to know the velocity components and just before impact. At the instant of release, the empty fuel tank has the same velocity as that of the plane. Therefore, the magnitudes of the initial velocity components of the fuel tank are given by and , where is the speed of the plane at the instant of release. Since the motion has zero acceleration, the component of the velocity of the plane remains equal to for all later times while the tank is airborne. The component of the velocity of the tank after it has undergone a vertical displacement is given by Equation 3.6b.
SOLUTION
(a)��
Taking up as the positive direction, the velocity components of the fuel tank just before it hits the ground are
From Equation 3.6b, we have
Therefore, the magnitude of the velocity of the fuel tank just before impact is
The velocity vector just before impact is inclined at an angle with respect to the horizontal. This angle is
(b)��
As shown in Conceptual Example 10, once the fuel tank in part a rises and falls to the same altitude at which it was released, its motion is identical to the fuel tank in part b. Therefore, the velocity of the fuel tank in part b just before impact is .
24.��
�A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of , hoping to land on the roof of an adjacent building. Air resistance is negligible. The horizontal distance between the two buildings is D, and the roof of the adjacent building is 2.0 m below the jumping-off point. Find the maximum value for D.
25.��
On a spacecraft, two engines are turned on for 684 s at a moment when the velocity of the craft has x and y components of and . While the engines are firing, the craft undergoes a displacement that has components of and . Find the x and y components of the craft's acceleration.
Answer:
26.��
�In the absence of air resistance, a projectile is launched from and returns to ground level. It follows a trajectory similar to that shown in Figure 3.10 and has a range of 23 m. Suppose the launch speed is doubled, and the projectile is fired at the same angle above the ground. What is the new range?
27.��
�A fire hose ejects a stream of water at an angle of above the horizontal. The water leaves the nozzle with a speed of . Assuming that the water behaves like a projectile, how far from a building should the fire hose be located to hit the highest possible fire?
Answer:
30.0 m
REASONING AND SOLUTION The water exhibits projectile motion. The component of the motion has zero acceleration while the component is subject to the acceleration due to gravity. In order to reach the highest possible fire, the displacement of the hose from the building is , where, according to Equation 3.5a (with ),
with equal to the time required for the water the reach its maximum vertical displacement. The time can be found by considering the vertical motion. From Equation 3.3b,
When the water has reached its maximum vertical displacement, . Taking up and to the right as the positive directions, we find that
and
Therefore, we have
28.��
Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of parallel to the ground. Upon contact with the bat the ball is 1.2 m above the ground. Player B wishes to duplicate this bunt, in so far as he also wants to give the ball a velocity parallel to the ground and have his ball travel the same horizontal distance as player A's ball does. However, player B hits the ball when it is 1.5 m above the ground. What is the magnitude of the initial velocity that player B's ball must be given?
29.��
A major-league pitcher can throw a baseball in excess of . If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0 m away from the point of release?
Answer:
0.844 m
30.��
A quarterback claims that he can throw the football a horizontal distance of 183 m (200 yd). Furthermore, he claims that he can do this by launching the ball at the relatively low angle of above the horizontal. To evaluate this claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. For comparison, a baseball pitcher who can accurately throw a fastball at (100 mph) would be considered exceptional.
31.��
�An eagle is flying horizontally at with a fish in its claws. It accidentally drops the fish.
(a)��
How much time passes before the fish's speed doubles?
Answer:
1.1 s
REASONING The speed of the fish at any time is given by , where and are the and components of the velocity at that instant. Since the horizontal motion of the fish has zero acceleration, for all times . Since the fish is dropped by the eagle, is equal to the horizontal speed of the eagle and . The component of the velocity of the fish for any time is given by Equation 3.3b with . Thus, the speed at any time t is given by .
SOLUTION
(a)��
The initial speed of the fish is . When the fish's speed doubles,. Therefore,
Assuming that downward is positive and solving for , we have
(b)��
When the fish's speed doubles again,. Therefore,
Solving for , we have
Therefore, the additional time for the speed to double again is .
(b)��
How much additional time would be required for the fish's speed to double again?
Answer:
1.3 s
REASONING The speed of the fish at any time is given by , where and are the and components of the velocity at that instant. Since the horizontal motion of the fish has zero acceleration, for all times . Since the fish is dropped by the eagle, is equal to the horizontal speed of the eagle and . The component of the velocity of the fish for any time is given by Equation 3.3b with . Thus, the speed at any time t is given by .
SOLUTION
(a)��
The initial speed of the fish is . When the fish's speed doubles,. Therefore,
Assuming that downward is positive and solving for , we have
(b)��
When the fish's speed doubles again,. Therefore,
Solving for , we have
Therefore, the additional time for the speed to double again is .
REASONING The speed of the fish at any time is given by , where and are the and components of the velocity at that instant. Since the horizontal motion of the fish has zero acceleration, for all times . Since the fish is dropped by the eagle, is equal to the horizontal speed of the eagle and . The component of the velocity of the fish for any time is given by Equation 3.3b with . Thus, the speed at any time t is given by .
SOLUTION
(a)��
The initial speed of the fish is . When the fish's speed doubles,. Therefore,
Assuming that downward is positive and solving for , we have
(b)��
When the fish's speed doubles again,. Therefore,
Solving for , we have
Therefore, the additional time for the speed to double again is .
32.��
�The perspective provided by Multiple-Concept Example 9 is useful here. The highest barrier that a projectile can clear is 13.5 m, when the projectile is launched at an angle of above the horizontal. What is the projectile's launch speed?
33.��
Consult Multiple-Concept Example 4 for background before beginning this problem. Suppose the water at the top of Niagara Falls has a horizontal speed of just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a angle below the horizontal?
Answer:
5.2 m
34.��
�On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 3.5 times as far as he would have on earth, given the same initial velocities on both planets. The ball is launched at a speed of at an angle of above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet, what are
(a)��
the maximum height and
(b)��
the range of the ball?
35.��
A rocket is fired at a speed of from ground level, at an angle of above the horizontal. The rocket is fired toward an 11.0-m-high wall, which is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?
Answer:
33.2 m
36.��
A rifle is used to shoot twice at a target, using identical cartridges. The first time, the rifle is aimed parallel to the ground and directly at the center of the bull's-eye. The bullet strikes the target at a distance of below the center, however. The second time, the rifle is similarly aimed, but from twice the distance from the target. This time the bullet strikes the target at a distance of below the center. Find the ratio .
*37.��
�An airplane with a speed of is climbing upward at an angle of with respect to the horizontal. When the plane's altitude is 732 m, the pilot releases a package.
(a)��
Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.
Answer:
1380 m
REASONING
(a)��
The drawing shows the initial velocity of the package when it is released. The initial speed of the package is . The component of its displacement along the ground is labeled as . The data for the x direction are indicated in the data table below.
t0012-nu
x-Direction Data
?
Since only two variables are known, it is not possible to determine from the data in this table. A value for a third variable is needed. We know that the time of flight is the same for both the and motions, so let's now look at the data in the direction.
y-Direction Data
?
Note that the displacement of the package points from its initial position toward the ground, so its value is negative, i.e., . The data in this table, along with the appropriate equation of kinematics, can be used to find the time of flight . This value for can, in turn, be used in conjunction with the direction data to determine .
(b)��
The drawing at the right shows the velocity of the package just before impact. The angle that the velocity makes with respect to the ground can be found from the inverse tangent function as . Once the time has been found in part (a), the values of and can be determined from the data in the tables and the appropriate equations of kinematics.
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SOLUTION
(a)��
To determine the time that the package is in the air, we will use Equation 3.5b and the data in the y-direction data table. Solving this quadratic equation for the time yields
We discard the first solution, since it is a negative value and, hence, unrealistic. The displacement can be found using , the data in the -direction data table, and Equation 3.5a:
(b)��
The angle that the velocity of the package makes with respect to the ground is given by . Since there is no acceleration in the direction , is the same as , so that . Equation 3.3b can be employed with the -direction data to find :
Therefore,
where the minus sign indicates that the angle is .
(b)��
Relative to the ground, determine the angle of the velocity vector of the package just before impact.
Answer:
below the horizontal
REASONING
(a)��
The drawing shows the initial velocity of the package when it is released. The initial speed of the package is . The component of its displacement along the ground is labeled as . The data for the x direction are indicated in the data table below.
t0012-nu
x-Direction Data
?
Since only two variables are known, it is not possible to determine from the data in this table. A value for a third variable is needed. We know that the time of flight is the same for both the and motions, so let's now look at the data in the direction.
y-Direction Data
?
Note that the displacement of the package points from its initial position toward the ground, so its value is negative, i.e., . The data in this table, along with the appropriate equation of kinematics, can be used to find the time of flight . This value for can, in turn, be used in conjunction with the direction data to determine .
(b)��
The drawing at the right shows the velocity of the package just before impact. The angle that the velocity makes with respect to the ground can be found from the inverse tangent function as . Once the time has been found in part (a), the values of and can be determined from the data in the tables and the appropriate equations of kinematics.
t0013-nu
SOLUTION
(a)��
To determine the time that the package is in the air, we will use Equation 3.5b and the data in the y-direction data table. Solving this quadratic equation for the time yields
We discard the first solution, since it is a negative value and, hence, unrealistic. The displacement can be found using , the data in the -direction data table, and Equation 3.5a:
(b)��
The angle that the velocity of the package makes with respect to the ground is given by . Since there is no acceleration in the direction , is the same as , so that . Equation 3.3b can be employed with the -direction data to find :
Therefore,
where the minus sign indicates that the angle is .
REASONING
(a)��
The drawing shows the initial velocity of the package when it is released. The initial speed of the package is . The component of its displacement along the ground is labeled as . The data for the x direction are indicated in the data table below.
t0012-nu
x-Direction Data
?
Since only two variables are known, it is not possible to determine from the data in this table. A value for a third variable is needed. We know that the time of flight is the same for both the and motions, so let's now look at the data in the direction.
y-Direction Data
?
Note that the displacement of the package points from its initial position toward the ground, so its value is negative, i.e., . The data in this table, along with the appropriate equation of kinematics, can be used to find the time of flight . This value for can, in turn, be used in conjunction with the direction data to determine .
(b)��
The drawing at the right shows the velocity of the package just before impact. The angle that the velocity makes with respect to the ground can be found from the inverse tangent function as . Once the time has been found in part (a), the values of and can be determined from the data in the tables and the appropriate equations of kinematics.
t0013-nu
SOLUTION
(a)��
To determine the time that the package is in the air, we will use Equation 3.5b and the data in the y-direction data table. Solving this quadratic equation for the time yields
We discard the first solution, since it is a negative value and, hence, unrealistic. The displacement can be found using , the data in the -direction data table, and Equation 3.5a:
(b)��
The angle that the velocity of the package makes with respect to the ground is given by . Since there is no acceleration in the direction , is the same as , so that . Equation 3.3b can be employed with the -direction data to find :
Therefore,
where the minus sign indicates that the angle is .
*38.��
�Multiple-Concept Example 4 deals with a situation similar to that presented here. A marble is thrown horizontally with a speed of from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of with the horizontal. From what height above the ground was the marble thrown?
*39.��
�Review Conceptual Example 5 before beginning this problem. You are traveling in a convertible with the top down. The car is moving at a constant velocity of , due east along flat ground. You throw a tomato straight upward at a speed of . How far has the car moved when you get a chance to catch the tomato?
Answer:
56 m
*40.��
�See Multiple-Concept Example 9 for the basic idea behind problems such as this. A diver springs upward from a diving board. At the instant she contacts the water, her speed is , and her body is extended at an angle of with respect to the horizontal surface of the water. At this instant her vertical displacement is , where downward is the negative direction. Determine her initial velocity, both magnitude and direction.
*41.��
�A soccer player kicks the ball toward a goal that is 16.8 m in front of him. The ball leaves his foot at a speed of and an angle of above the ground. Find the speed of the ball when the goalie catches it in front of the net.
Answer:
*42.��
�In the javelin throw at a track-and-field event, the javelin is launched at a speed of at an angle of above the horizontal. As the javelin travels upward, its velocity points above the horizontal at an angle that decreases as time passes. How much time is required for the angle to be reduced from at launch to ?
*43.��
�An airplane is flying with a velocity of at an angle of with the horizontal, as the drawing shows. When the altitude of the plane is 2.4 km, a flare is released from the plane. The flare hits the target on the ground. What is the angle ?
image
Answer:
REASONING The angle can be found from
��(1)
where is the horizontal displacement of the flare. Since , it follows that . The flight time is determined by the vertical motion. In particular, the time t can be found from Equation 3.5b. Once the time is known, can be calculated.
SOLUTION From Equation 3.5b, assuming upward is the positive direction, we have
which can be rearranged to give the following equation that is quadratic in t:
Using and and suppressing the units, we obtain the quadratic equation
Using the quadratic formula, we obtain . Therefore, we find that
Equation 1 then gives
*44.��
�A child operating a radio-controlled model car on a dock accidentally steers it off the edge. The car's displacement 1.1 s after leaving the dock has a magnitude of 7.0 m. What is the car's speed at the instant it drives off the edge of the dock?
*45.��
�After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 51.0 m horizontally from the end of the ramp. His velocity, just before landing, is and points in a direction below the horizontal. Neglecting air resistance and any lift he experiences while airborne, find his initial velocity (magnitude and direction) when he left the end of the ramp. Express the direction as an angle relative to the horizontal.
Answer:
, above the horizontal
*46.��
�Stones are thrown horizontally with the same velocity from the tops of two different buildings. One stone lands twice as far from the base of the building from which it was thrown as does the other stone. Find the ratio of the height of the taller building to the height of the shorter building.
**47.��
�The drawing shows an exaggerated view of a rifle that has been “sighted in” for a 91.4-meter target. If the muzzle speed of the bullet is , what are the two possible angles and between the rifle barrel and the horizontal such that the bullet will hit the target? One of these angles is so large that it is never used in target shooting.
image
Problem 47
(Hint: The following trigonometric identity may be useful: .)
Answer:
and
REASONING AND SOLUTION In the absence of air resistance, the bullet exhibits projectile motion. The component of the motion has zero acceleration while the component of the motion is subject to the acceleration due to gravity. The horizontal distance traveled by the bullet is given by Equation 3.5a (with ):
with equal to the time required for the bullet to reach the target. The time can be found by considering the vertical motion. From Equation 3.3b,
When the bullet reaches the target, . Assuming that up and to the right are the positive directions, we have
Using the fact that , we have
Thus, we find that
and
Therefore,
**48.��
�A projectile is launched from ground level at an angle of above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?
**49.��
�From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of , parallel to the ground. As the drawing shows, the bullet puts a hole in a window of another building and hits the wall that faces the window. Using the data in the drawing, determine the distances D and H, which locate the point where the gun was fired. Assume that the bullet does not slow down as it passes through the window.
image
Answer:
REASONING Since the horizontal motion is not accelerated, we know that the component of the velocity remains constant at 340 m/s. Thus, we can use Equation 3.5a (with ) to determine the time that the bullet spends in the building before it is embedded in the wall. Since we know the vertical displacement of the bullet after it enters the building, we can use the flight time in the building and Equation 3.5b to find the component of the velocity of the bullet as it enters the window. Then, Equation 3.6b can be used (with ) to determine the vertical displacement of the bullet as it passes between the buildings. We can determine the distance by adding the magnitude of to the vertical distance of 0.50 m within the building.
Once we know the vertical displacement of the bullet as it passes between the buildings, we can determine the time required for the bullet to reach the window using Equation 3.4b. Since the motion in the direction is not accelerated, the distance D can then be found from .
SOLUTION Assuming that the direction to the right is positive, we find that the time that the bullet spends in the building is (according to Equation 3.5a)
The vertical displacement of the bullet after it enters the building is, taking down as the negative direction, equal to . Therefore, the vertical component of the velocity of the bullet as it passes through the window is, from Equation 3.5b,
The vertical displacement of the bullet as it travels between the buildings is (according to Equation 3.6b with )
Therefore, the distance is
The time for the bullet to reach the window, according to Equation 3.4b, is
Hence, the distance is given by
**50.��
In the annual battle of the dorms, students gather on the roofs of Jackson and Walton dorms to launch water balloons at each other with slingshots. The horizontal distance between the buildings is 35.0 m, and the heights of the Jackson and Walton buildings are, respectively, 15.0 m and 22.0 m. Ignore air resistance.
(a)��
The first balloon launched by the Jackson team hits Walton dorm 2.0 s after launch, striking it halfway between the ground and the roof. Find the direction of the balloon's initial velocity. Give your answer as an angle measured above the horizontal.
(b)��
A second balloon launched at the same angle hits the edge of Walton's roof. Find the initial speed of this second balloon.
**51.��
Two cannons are mounted as shown in the drawing and rigged to fire simultaneously. They are used in a circus act in which two clowns serve as human cannonballs. The clowns are fired toward each other and collide at a height of 1.00 m above the muzzles of the cannons. Clown A is launched at a angle, with a speed of . The horizontal separation between the clowns as they leave the cannons is 6.00 m. Find the launch speed and the launch angle for clown B.
image
Answer:
,
Section 3.4 Relative Velocity
52.��
In a marathon race Chad is out in front, running due north at a speed of . John is 95 m behind him, running due north at a speed of . How long does it take for John to pass Chad?
53.��
�A swimmer, capable of swimming at a speed of in still water (i.e., the swimmer can swim with a speed of relative to the water), starts to swim directly across a 2.8-km-wide river. However, the current is , and it carries the swimmer downstream.
(a)��
How long does it take the swimmer to cross the river?
Answer:
REASONING The velocity of the swimmer relative to the ground is the vector sum of the velocity of the swimmer relative to the water and the velocity of the water relative to the ground as shown at the right: .
t0020-nu
The component of that is parallel to the width of the river determines how fast the swimmer is moving across the river; this parallel component is . The time for the swimmer to cross the river is equal to the width of the river divided by the magnitude of this velocity component.
The component of that is parallel to the direction of the current determines how far the swimmer is carried down stream; this component is . Since the motion occurs with constant velocity, the distance that the swimmer is carried downstream while crossing the river is equal to the magnitude of multiplied by the time it takes for the swimmer to cross the river.
SOLUTION
(a)��
The time for the swimmer to cross the river is
(b)��
The distance that the swimmer is carried downstream while crossing the river is
(b)��
How far downstream will the swimmer be upon reaching the other side of the river?
Answer:
REASONING The velocity of the swimmer relative to the ground is the vector sum of the velocity of the swimmer relative to the water and the velocity of the water relative to the ground as shown at the right: .
t0020-nu
The component of that is parallel to the width of the river determines how fast the swimmer is moving across the river; this parallel component is . The time for the swimmer to cross the river is equal to the width of the river divided by the magnitude of this velocity component.
The component of that is parallel to the direction of the current determines how far the swimmer is carried down stream; this component is . Since the motion occurs with constant velocity, the distance that the swimmer is carried downstream while crossing the river is equal to the magnitude of multiplied by the time it takes for the swimmer to cross the river.
SOLUTION
(a)��
The time for the swimmer to cross the river is
(b)��
The distance that the swimmer is carried downstream while crossing the river is
REASONING The velocity of the swimmer relative to the ground is the vector sum of the velocity of the swimmer relative to the water and the velocity of the water relative to the ground as shown at the right: .
t0020-nu
The component of that is parallel to the width of the river determines how fast the swimmer is moving across the river; this parallel component is . The time for the swimmer to cross the river is equal to the width of the river divided by the magnitude of this velocity component.
The component of that is parallel to the direction of the current determines how far the swimmer is carried down stream; this component is . Since the motion occurs with constant velocity, the distance that the swimmer is carried downstream while crossing the river is equal to the magnitude of multiplied by the time it takes for the swimmer to cross the river.
SOLUTION
(a)��
The time for the swimmer to cross the river is
(b)��
The distance that the swimmer is carried downstream while crossing the river is
54.��
�Two friends, Barbara and Neil, are out rollerblading. With respect to the ground, Barbara is skating due south at a speed of . Neil is in front of her. With respect to the ground, Neil is skating due west at a speed of . Find Neil's velocity (magnitude and direction relative to due west), as seen by Barbara.
55.��
A police officer is driving due north at a constant speed of relative to the ground when she notices a truck on an east–west highway ahead of her, driving west at high speed. She finds that the truck's speed relative to her car is (about 110 mph).
(a)��
Sketch the vector triangle that shows how the truck's velocity relative to the ground is related to the police car's velocity relative to the ground and to the truck's velocity relative to the police car. The sketch need not be to scale, but the velocity vectors should be oriented correctly and bear the appropriate labels.
Answer:
The answer is a drawing.
(b)��
What is the truck's speed, relative to the ground?
Answer:
56.��
At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt on which you can either stand or walk. Suppose a speed ramp has a length of 105 m and is moving at a speed of relative to the ground. In addition, suppose you can cover this distance in 75 s when walking on the ground. If you walk at the same rate with respect to the speed ramp that you walk on the ground, how long does it take for you to travel the 105 m using the speed ramp?
57.��
You are in a hot-air balloon that, relative to the ground, has a velocity of in a direction due east. You see a hawk moving directly away from the balloon in a direction due north. The speed of the hawk relative to you is . What are the magnitude and direction of the hawk's velocity relative to the ground? Express the directional angle relative to due east.
Answer:
, north of east
58.��
�On a pleasure cruise a boat is traveling relative to the water at a speed of due south. Relative to the boat, a passenger walks toward the back of the boat at a speed of .
(a)��
What are the magnitude and direction of the passenger's velocity relative to the water?
(b)��
How long does it take for the passenger to walk a distance of 27 m on the boat?
(c)��
How long does it take for the passenger to cover a distance of 27 m on the water?
59.��
�Two passenger trains are passing each other on adjacent tracks. Train A is moving east with a speed of , and train B is traveling west with a speed of .
(a)��
What is the velocity (magnitude and direction) of train A as seen by the passengers in train B?
Answer:
due east
REASONING The velocity of train A relative to train B is the vector sum of the velocity of train A relative to the ground and the velocity of the ground relative to train B, as indicated by Equation 3.7: . The values of and are given in the statement of the problem. We must also make use of the fact that .
SOLUTION
(a)��
Taking east as the positive direction, the velocity of A relative to B is, according to Equation 3.7,
The positive sign indicates that the direction of is .
(b)��
Similarly, the velocity of B relative to A is
The negative sign indicates that the direction of is .
(b)��
What is the velocity (magnitude and direction) of train B as seen by the passengers in train A?
Answer:
due west
REASONING The velocity of train A relative to train B is the vector sum of the velocity of train A relative to the ground and the velocity of the ground relative to train B, as indicated by Equation 3.7: . The values of and are given in the statement of the problem. We must also make use of the fact that .
SOLUTION
(a)��
Taking east as the positive direction, the velocity of A relative to B is, according to Equation 3.7,
The positive sign indicates that the direction of is .
(b)��
Similarly, the velocity of B relative to A is
The negative sign indicates that the direction of is .
REASONING The velocity of train A relative to train B is the vector sum of the velocity of train A relative to the ground and the velocity of the ground relative to train B, as indicated by Equation 3.7: . The values of and are given in the statement of the problem. We must also make use of the fact that .
SOLUTION
(a)��
Taking east as the positive direction, the velocity of A relative to B is, according to Equation 3.7,
The positive sign indicates that the direction of is .
(b)��
Similarly, the velocity of B relative to A is
The negative sign indicates that the direction of is .
60.��
�The captain of a plane wishes to proceed due west. The cruising speed of the plane is relative to the air. A weather report indicates that a 38.0-m/s wind is blowing from the south to the north. In what direction, measured with respect to due west, should the pilot head the plane?
*61.��
�A person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of relative to the ground. When the train moves at a constant velocity, the raindrops make an angle of when they move past the window, as the drawing shows. How fast is the train moving?
image
Answer:
*62.��
�A ferryboat is traveling in a direction north of east with a speed of relative to the water. A passenger is walking with a velocity of due east relative to the boat. What is the velocity (magnitude and direction) of the passenger with respect to the water? Determine the directional angle relative to due east.
*63.��
�Mario, a hockey player, is skating due south at a speed of relative to the ice. A teammate passes the puck to him. The puck has a speed of and is moving in a direction of west of south, relative to the ice. What are the magnitude and direction (relative to due south) of the puck's velocity, as observed by Mario?
Answer:
, west of south
REASONING The velocity of the puck relative to Mario is the vector sum of the velocity of the puck relative to the ice and the velocity of the ice relative to Mario as indicated by Equation 3.7: . The values of and are given in the statement of the problem. In order to use the data, we must make use of the fact that , with the result that .
SOLUTION The first two rows of the following table give the east/west and north/south components of the vectors and . The third row gives the components of their resultant . Due east and due north have been taken as positive.
Vector
East/West Component
North/South Component
0
Now that the components of are known, the Pythagorean theorem can be used to find the magnitude.
t0027-nu
The direction of is found from
**64.��
A jetliner can fly 6.00 hours on a full load of fuel. Without any wind it flies at a speed of . The plane is to make a round-trip by heading due west for a certain distance, turning around, and then heading due east for the return trip. During the entire flight, however, the plane encounters a 57.8-m/s wind from the jet stream, which blows from west to east. What is the maximum distance that the plane can travel due west and just be able to return home?
**65.��
�Two boats are heading away from shore. Boat 1 heads due north at a speed of relative to the shore. Relative to boat 1, boat 2 is moving north of east at a speed of . A passenger on boat 2 walks due east across the deck at a speed of relative to boat 2. What is the speed of the passenger relative to the shore?
Answer:
REASONING The relative velocities in this problem are:
= velocity of the Passenger relative to the Shore
= velocity of the Passenger relative to Boat 2 (1.20 m/s, due east)
= velocity of Boat 2 relative to the Shore
= velocity of Boat 2 relative to Boat 1 (1.60 m/s, at north of east)
= velocity of Boat 1 relative to the Shore (3.00 m/s, due north)
The velocity of the passenger relative to the shore is related to and by (see the method of subscripting discussed in Section 3.4):
But , the velocity of Boat 2 relative to the shore, is related to and by
t0028-nu
Substituting this expression for into the first equation yields
This vector sum is shown in the diagram. We will determine the magnitude of from the equation above by using the method of scalar components.
SOLUTION The table below lists the scalar components of the four vectors in the drawing.
Vector
x Component
y Component
0 m/s
0 m/s
The magnitude of can be found by applying the Pythagorean theorem to its and components:
�� Additional Problems
66.��
Useful background for this problem can be found in Multiple-Concept Example 2. On a spacecraft two engines fire for a time of 565 s. One gives the craft an acceleration in the x direction of , while the other produces an acceleration in the y direction of . At the end of the firing period, the craft has velocity components of and . Find the magnitude and direction of the initial velocity. Express the direction as an angle with respect to the axis.
67.��
A dolphin leaps out of the water at an angle of above the horizontal. The horizontal component of the dolphin's velocity is . Find the magnitude of the vertical component of the velocity.
Answer:
68.��
A hot-air balloon is rising straight up with a speed of . A ballast bag is released from rest relative to the balloon at 9.5 m above the ground. How much time elapses before the ballast bag hits the ground?
69.��
A golf ball rolls off a horizontal cliff with an initial speed of . The ball falls a vertical distance of 15.5 m into a lake below.
(a)��
How much time does the ball spend in the air?
Answer:
1.78 s
(b)��
What is the speed v of the ball just before it strikes the water?
Answer:
70.��
�When chasing a hare along a flat stretch of ground, a greyhound leaps into the air at a speed of , at an angle of above the horizontal.
(a)��
What is the range of his leap and
(b)��
for how much time is he in the air?
71.��
�Multiple-Concept Example 4 provides useful background for this problem. A diver runs horizontally with a speed of off a platform that is 10.0 m above the water. What is his speed just before striking the water?
Answer:
REASONING Once the diver is airborne, he moves in the direction with constant velocity while his motion in the direction is accelerated (at the acceleration due to gravity). Therefore, the magnitude of the component of his velocity remains constant at 1.20 m/s for all times . The magnitude of the component of the diver's velocity after he has fallen through a vertical displacement can be determined from Equation 3.6b: . Since the diver runs off the platform horizontally, . Once the and components of the velocity are known for a particular vertical displacement , the speed of the diver can be obtained from .
SOLUTION For convenience, we will take downward as the positive direction. After the diver has fallen 10.0 m, the component of his velocity is, from Equation 3.6b,
Therefore,
72.��
�A ball is thrown upward at a speed at an angle of above the horizontal. It reaches a maximum height of 7.5 m. How high would this ball go if it were thrown straight upward at speed ?
*73.��
�A golfer, standing on a fairway, hits a shot to a green that is elevated 5.50 m above the point where she is standing. If the ball leaves her club with a velocity of at an angle of above the ground, find the time that the ball is in the air before it hits the green.
Answer:
5.17 s
*74.��
�In a stunt being filmed for a movie, a sports car overtakes a truck towing a ramp, drives up and off the ramp, soars into the air, and then lands on top of a flat trailer being towed by a second truck. The tops of the ramp and the flat trailer are the same height above the road, and the ramp is inclined above the horizontal. Both trucks are driving at a constant speed of , and the flat trailer is 15 m from the end of the ramp. Neglect air resistance, and assume that the ramp changes the direction, but not the magnitude, of the car's initial velocity. What is the minimum speed the car must have, relative to the road, as it starts up the ramp?
*75.��
�As preparation for this problem, review Conceptual Example 10. The two stones described there have identical initial speeds of and are thrown at an angle , one below the horizontal and one above the horizontal. What is the distance between the points where the stones strike the ground?
Answer:
14.9 m
REASONING The horizontal distance covered by stone 1 is equal to the distance covered by stone 2 after it passes point P in the following diagram. Thus, the distance between the points where the stones strike the ground is equal to , the horizontal distance covered by stone 2 when it reaches P. In the diagram, we assume up and to the right are positive.
t0030-nu
SOLUTION If is the time required for stone 2 to reach P, then
For the vertical motion of stone 2, . Solving for gives
When stone 2 reaches P, , so the time required to reach P is
Then,
*76.��
�Relative to the ground, a car has a velocity of , directed due north. Relative to this car, a truck has a velocity of , directed north of east. What is the magnitude of the truck's velocity relative to the ground?
*77.��
�The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his head, forcing him to move quickly away from the net (see the drawing). Suppose that you lob the ball with an initial speed of , at an angle of above the horizontal. At this instant your opponent is 10.0 m away from the ball. He begins moving away from you 0.30 s later, hoping to reach the ball and hit it back at the moment that it is 2.10 m above its launch point. With what minimum average speed must he move? (Ignore the fact that he can stretch, so that his racket can reach the ball before he does.)
image
Answer:
**78.��
A Coast Guard ship is traveling at a constant velocity of , due east, relative to the water. On his radar screen the navigator detects an object that is moving at a constant velocity. The object is located at a distance of 2310 m with respect to the ship, in a direction south of east. Six minutes later, he notes that the object's position relative to the ship has changed to 1120 m, south of west. What are the magnitude and direction of the velocity of the object relative to the water? Express the direction as an angle with respect to due west.
**79.��
A placekicker is about to kick a field goal. The ball is 26.9 m from the goalpost. The ball is kicked with an initial velocity of at an angle above the ground. Between what two angles, and , will the ball clear the 2.74-m-high crossbar?
(Hint: The following trigonometric identities may be useful: and .)
Answer:
and


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