## Conceptual Physics (12th Edition)

a. The velocity of the ball at its highest point is zero, for just a moment. b. One second before reaching the top, the ball has a velocity of 10 m/s. c. The change in its velocity is 0-10 m/s = -10 m/s for this 1-second interval (or for any other 1-second-long interval during freefall). d. One second after reaching the top, its velocity is -10 m/s, which equal in magnitude but directed opposite to its velocity 1 second before reached the highest point. e. The change in velocity is -10 m/s - 0 m/s = -10 m/s for this 1-second interval (or for any other 1-second-long interval during freefall). f. For the 2-second interval, the change in velocity is -10 m/s - 10 m/s = -20 m/s. A common mistake is to think that the change is zero, but there is a nonzero change in velocity because the ball's direction has changed. g. The acceleration of the ball is $-10 \frac{m}{s^{2}}$ at all times. The magnitude of freefall acceleration is $g = 10 \frac{m}{s^{2}}$, and a minus sign is typically used to indicate that the acceleration is directed downward, toward the center of the Earth.