College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 7 - Multiple-Choice Questions - Page 261: 12


The correct answer is: (e)

Work Step by Step

The golf ball begins its upward climb with a kinetic energy of $\frac{1}{2}mv_0^2$. At the ground, the initial gravitational potential energy is zero. As the height increases, the kinetic energy is gradually converted into gravitational potential energy until the gravitational potential energy reaches its maximum value and the kinetic energy is zero when the ball is at maximum height. On the way down, the kinetic energy increases as the gravitational potential energy is converted back into kinetic energy. Let the initial velocity be $v_0$. We can find an expression for the velocity as a function of time: $v = v_0-9.80~t$ We can find an expression for the kinetic energy as a function of time: $KE = \frac{1}{2}mv^2$ $KE = \frac{1}{2}m(v_0-9.80~t)^2$ $KE = \frac{1}{2}m~(94.04~t^2-19.60~v_0~t+v_0^2)$ This is the equation of a parabola that is concave up. Note that $KE = \frac{1}{2}mv_0^2$ at time $t = 0$. The correct answer is: (e)
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