## Conceptual Physics (12th Edition)

a. As far as vertical motion is concerned, the tennis ball starts from rest, and drops y meters. Relate the height dropped to the time. $$y = \frac{1}{2}g t^{2}$$ $$t = \sqrt{\frac{2 y}{g}}$$ This is discussed on page 48. The horizontal velocity of the tennis ball does not change. At maximum speed, the horizontal distance it moves is d. $$d = vt = (v)\sqrt{\frac{2 y}{g}}$$ Solve for the maximum speed. $$v = \frac{d}{t} = \frac{d}{\sqrt{\frac{2 y}{g}}}$$ b. Plug the numbers given into the equation just derived to arrive at v = 26.8 m/s. c. No. The ball is in free fall and the acceleration is independent of mass.