Answer
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) $cos^2~\theta$.
