First state the givens:
Now we can find the velocity components:
Using the x component of velocity we can find the time it takes for the ball to travel through the air:
Given the time of flight we can now find the height of the building. Consider the ground level to be

y = 0,

and the positive y direction to be upwards.
The height of the building is 110 m.

Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values.

The free body diagram for each system is shown below.

N

is the normal force exerted by the surface on the system and

W

refers to the weight. Note that for the system containing both boxes, the contact force

F_C

is not external to the system. And when you consider just box B, the force

F_p

does not act on it.

(a) To find the acceleration of the two boxes, consider both boxes as the system. Apply Newton's second law

F_{net, x} = ma_x

for motion in the horizontal direction to this system, noting that the only external force acting on this system is the pushing force. Since all of the forces are along the x axis, we can drop the vector notation.
(b) To find the magnitude of the contact force

F_C

exerted on box B by box A, consider box B as the system. In this case, the net force acting on the system is just the contact force.
(c) If Alex now pushed on box B rather than on box A, you would merely have to switch the two masses in the solutions above. Since the pushing force is the same as in part (a), the two boxes would move with the same acceleration. Therefore the contact force would now be larger since mass of A is larger. We can check this out by performing the calculations.

Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values.

(a) The diagram shows the components of the forces in the x and z directions. We can write the two forces in terms of the unit vectors î and . The net force will be the vector sum of these two forces.
(b) We now apply Newton's second law where

F_net

is the net force acting on the mass m and

a

is the acceleration. Solving for

a,

we get

(a) The free-body diagram for the system (the block of ice) is shown below. Here

N

is the normal force exerted by the lake surface on the block of ice. Note that the applied force vector

F

is drawn with its tail at the system to easily visualize its components. Since we are interested in determining the magnitude of the normal force, we need to consider motion in the vertical direction when applying Newton's second law. Note that the acceleration in the vertical direction is zero as there is no motion in that direction.

Newton's second law for the vertical direction can therefore be written as follows.
(b) In this case, the pulling force is pointing up and to the right. The sine component of

F

is therefore directed upward now. As in part (a), the acceleration in the vertical direction is zero and Newton's second law can be written as follows.

(a) The diagram shows the forces, the velocity, and acceleration vectors for the car when it is at A. The centripetal force acting on the car is given by the equation where v is the speed at A and r is the radius of the curved path. Note that centripetal force is always directed toward the center of the circular path.

The centripetal force acting on the car is the vector sum of the weight of the car and the normal force of the track on the car. Taking +y to be up, we can write this as Combining Equations (1) and (2) and solving for N, we get Substitute the given values to find N. The diagram shows the forces, the velocity, and acceleration vectors for the car when it is at B. Note that the centripetal force, which is always directed toward the center of the circular path, is now in the upward direction. Equations (2) and (3) now become Solve for N and substitute the known values for the car when it is at B. Substitute the given values to find N. Note that the normal force on the car is greater when it is at B.

(b) The free-body diagram for the car at C will be similar to that at A. The car will lose contact with the track when the normal force is zero. So, set

N = 0

in Equation (3) and solve for the speed. For speeds equal to and greater than 7.67 m/s, the car will lose contact with the track.

If we consider the system to be the three blocks and the connecting cords, the horizontal surface, and Earth, from conservation of energy we may write where K is the kinetic energy, U is the potential energy, and E_int is the internal energy of the system.

The change in kinetic energy of the three blocks may be expressed as where v is the speed of the blocks after they have traveled a distance d.

The change in gravitational potential energy of the three blocks may be expressed as The relationship between the change in internal energy, the friction force acting on m₂, and the distance m₂ slides along the horizontal surface may be expressed as Substituting the change in kinetic energy, change in potential energy, and change in internal energy into the above energy statement, obtain Solving for the speed, obtain Entering values, we have Since we have been asked for the speed, the positive value is given.