Answer
(a) $ v(t)=600 -9 t^{2} ~m/s$
(b) $24~m/s$,$ -300~m/s$
(c) $3266~m$
Work Step by Step
We are given $$s(t)=600 t-3 t^{3} \text { for } 0 \leq t \leq 12$$
(a) We take the derivative of position to find the velocity:
\begin{align*}
v(t)&=s'(t) \\
&= 600 -9 t^{2}
\end{align*}
See the plot below.
(b)
We plug in the $t$ values into velocity:
\begin{align*}
v(8)&= 600 -9 (8)^{2}=24~m/s\\
v(10)&= 600 -9 (10)^{2}=-300~m/s
\end{align*}
(c) The maximum height when $v=0$ is at $ t= \dfrac{10\sqrt{6}}{3}$, so we have
$$s( \dfrac{10\sqrt{6}}{3})= 3266~m$$
