Answer
Proof given below.
Work Step by Step
Define "up" as the positive direction.
Solving the preceding exercise, we found that
$v(t)=-32t+v_{0}\ \ \mathrm{f}\mathrm{t}/\mathrm{s}$
Velocity is the derivative of position, $v(t)=s'(t)$, so
$s(t)=\displaystyle \int(-32t+v_{0})dt=-16t^{2}+v_{0}t+C.$
If $h$ is the starting height, $ s(0)=h$
$h=0+0+C$
$C=h$
Thus,
$s(t)=-16t^{2}+v_{0}t+h$
The result of the preceding exercise is that the projectile reaches its highest point at
$t=\displaystyle \frac{v_{0}}{32}$
At this value of t, its height is
$s(\displaystyle \frac{v_{0}}{32})=-16(\frac{v_{0}}{32})^{2}+v_{0}(\frac{v_{0}}{32})+h$
$=\displaystyle \frac{-16v_{0}^{2}+32v_{0}^{2}}{32^{2}}+h$
$=\displaystyle \frac{16v_{0}^{2}}{32^{2}}+h$
$=\displaystyle \frac{v_{0}^{2}}{64}+h\ \ $ ft,
which is $\displaystyle \frac{v_{0}^{2}}{64}$ feet above the starting point.
