Answer
$\displaystyle \frac{v_{0}^{2}}{2g}+s_{0}$
Work Step by Step
At the point of maximum height, velocity is zero for a moment.
Velocity is the derivative of the position function
$v(t)=-gt+v_{0}$
We find the $t$ for which $v(t)=0$
$gt=v_{0}$
$t=\displaystyle \frac{v_{0}}{g}$
Evaluate $s(t)$ for this value
$s(\displaystyle \frac{v_{0}}{g})=-\frac{1}{2}g(\frac{v_{0}}{g})^{2}+v_{0}(\frac{v_{0}}{g})+s_{0}$
$=-\displaystyle \frac{v_{o}^{2}}{2g}+\frac{v_{0}^{2}}{g}+s_{0}$
$=\displaystyle \frac{-v_{0}+2v_{0}}{2g}+s_{0}$
= $\displaystyle \frac{v_{0}^{2}}{2g}+s_{0}$
