## College Physics (4th Edition)

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