Answer
Maximum height $s_0+\frac{v_0^2}{2g}$ meters when $t=\frac{v_0}{g}$ sec.
Work Step by Step
Step 1. Given the function $s=-\frac{1}{2}gt^2+v_0t+s_0$ where $s_0=s(0)$ is the starting height, we can find its derivative as $s'(t)=-gt+v_0$.
Step 2. The critical points can be found when $s'=0$ or undefined, which happens only when $t=\frac{v_0}{g}$, which gives $s=-\frac{1}{2}g(\frac{v_0}{g})^2+v_0(\frac{v_0}{g})+s_0=s_0+\frac{v_0^2}{2g}$ and we can identify this as a maximum (it can not be a minimum since $g\gt0$ and $\frac{v_0^2}{2g}\gt0$ ).
Step 3. We conclude that the body’s maximum height is $s_0+\frac{v_0^2}{2g}$ meters when $t=\frac{v_0}{g}$ sec.