Answer
$200 ~\mathrm{m} / \mathrm{s}$
$2040.82 \mathrm{~m}$
Work Step by Step
We have the position
\begin{align*}
s(t)&=s_{0}+v_{0} t-\frac{1}{2} g t^{2}\\
&=200 t-4.9 t^{2}\end{align*}
Then the velocity is (derivative of position):
$$v(t)=200-9.8 t $$
The maximum velocity of $200 \mathrm{m} / \mathrm{s}$ occurs at $t=0$.
The maximum height occurs when $v=0 $ at $t \approx 20.41\mathrm{s}$
Hence, the height is given by
$$s(20.41) =200 (20.41)-4.9 (20.41)^{2}\approx 2040.82 \mathrm{~m}$$
