Answer
$$19.6 \mathrm{m}$$
Work Step by Step
We have the position function
\begin{align*}
s(t)&=s_{0}+v_{0} t-\frac{1}{2} g t^{2}\\
&=v_{0} t-4.9 t^{2}\end{align*} When the ball hits the ground after $4$ seconds, its height is $0$: \begin{align*}
0&=s(4)\\
&=4 v_{0}-4.9(4)^{2}
\end{align*} then $v_{0}=19.6 \mathrm{m} / \mathrm{s} .$
The ball reaches its maximum height when $v=0,$
\begin{align*}
v&= v_0-9.8t\\
0&=19.6-9.8t\\
\end{align*}
Hence, $t=2$
At $t=2 $, , we have the height
\begin{align*}
s(2)&=0+19.6(2)-\frac{1}{2}(9.8)(4)\\
&=19.6 \mathrm{m}
\end{align*}
