College Physics (4th Edition)

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

Chapter 6 - Multiple-Choice Questions - Page 226: 11


The correct answer is: (f) $cos^2~\theta$

Work Step by Step

We can find an expression for the initial kinetic energy: $KE = \frac{1}{2}mv_0^2$ At the top of the trajectory, the speed of the projectile is $v_0~cos~\theta$, which is the horizontal component of the initial velocity. We can find an expression for the kinetic energy at the top of its trajectory: $KE = \frac{1}{2}m(v_0~cos~\theta)^2$ We can find the fraction of the initial kinetic energy which remains at the top of the trajectory: $\frac{\frac{1}{2}m(v_0~cos~\theta)^2 }{\frac{1}{2}mv_0^2} = cos^2~\theta$ The correct answer is: (f) $cos^2~\theta$.
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