Answer
(a) $s(t)=-16t^2+200t+50$
(b) $ 6.25\ sec$, $675\ ft$.
(c) $1.41\ sec \lt t \lt 11.09\ sec$
(d) $12.75\ sec$
Work Step by Step
(a) Known $s_0=50\ ft, v_0=200\ ft/sec$, we have the function as $s(t)=-16t^2+v_0t+s_2=-16t^2+200t+50$
(b) Rewrite the function as $s(t)=-16(t^2-\frac{25}{2}t+(\frac{25}{2})^2-(\frac{25}{2})^2)+50=-16(t-\frac{25}{4})^2+675$, thus the rocket reaches its maximum height at $t=\frac{25}{4}=6.25\ sec$ and the maximum height is $675\ ft$.
(c) Let $s(t)\gt 300$, we have $-16(t-\frac{25}{4})^2+675\gt 300$ or $(t-\frac{25}{4})^2\lt \frac{375}{16}$ which gives $1.41\ sec \lt t \lt 11.09\ sec$
(d) Let $s(t)=0$, we have $-16(t-\frac{25}{4})^2+675=0$ which gives $t\approx12.75\ sec$ (discard the negative answer)
