Question 8.1: Ball in Free Fall A ball of mass m is dropped from a height ......

(A) Conceptualize Active Figure 8.4 and our everyday experience with falling objects allow us to conceptualize the situation. Although we can readily solve this problem with the techniques of Chapter 2 , let us practice an energy approach.

Categorize We identify the system as the ball and the Earth. Because there is neither air resistance nor any other interaction between the system and the environment, the system is isolated and we use the isolated system model. The only force between members of the system is the gravitational force, which is conservative.

Analyze Because the system is isolated and there are no nonconservative forces acting within the system, we apply the principle of conservation of mechanical energy to the ball-Earth system. At the instant the ball is released, its kinetic energy is K_{i}=0 and the gravitational potential energy of the system is U_{g i}=m g h. When the ball is at a position y above the ground, its kinetic energy is K_{f}=\frac{1}{2} m v_{f}^{2} and the potential energy relative to the ground is U_{g f}=m g y.

Apply Equation 8.10:

\begin{aligned}& K_{f}+U_{g f}=K_{i}+U_{g i} \\& \frac{1}{2} m v_{f}^{2}+m g y=0+m g h\end{aligned}

Solve for v_{f} :

v_{f}^{2}=2 g(h-y) \rightarrow v_{f}=\sqrt{2 g(h-y)}

The speed is always positive. If you had been asked to find the ball’s velocity, you would use the negative value of the square root as the y component to indicate the downward motion.

(B) Analyze In this case, the initial energy includes kinetic energy equal to \frac{1}{2} m v_{i}{ }^{2}.

Apply Equation 8.10:

\frac{1}{2} m v_{f}^{2}+m g y=\frac{1}{2} m v_{i}^{2}+m g h

Solve for v_{f} :

v_{f}^{2}=v_{i}^{2}+2 g(h-y) \rightarrow v_{f}=\sqrt{v_{i}^{2}+2 g(h-y)}

Finalize This result for the final speed is consistent with the expression v_{y f}^{2}=v_{y i}^{2}-2 g\left(y_{f}-y_{i}\right) from the particle under constant acceleration model for a falling object, where y_{i}=h. Furthermore, this result is valid even if the initial velocity is at an angle to the horizontal (Quick Quiz 8.4) for two reasons: (1) the kinetic energy, a scalar, depends only on the magnitude of the velocity; and (2) the change in the gravitational potential energy of the system depends only on the change in position of the ball in the vertical direction.