Answer
(a) $-16t^2+90t$
(b) $2.81\ sec$, $126.56\ ft$.
(c) $2.17\ sec \lt t \lt 3.45\ sec$
(d) $5.63\ sec$
Work Step by Step
(a) Known $s_0=0\ ft, v_0=90\ ft/sec$, we have the function as $s(t)=-16t^2+v_0t+s_2=-16t^2+90t$
(b) Let $a=-16, b=90$, the rocket reaches its maximum height at $t=-\frac{b}{2a}=\frac{90}{32}\approx2.81\ sec$ and the maximum height is $s(2.81)\approx126.56\ ft$.
(c) Let $s(t)\gt 120$, we have $-16t^2+90t\gt 120$ or $16t^2-90t+120\lt 0$ which gives $2.17\ sec \lt t \lt 3.45\ sec$
(d) Let $s(t)=0$, we have $-16t^2+90t=0$ which gives $t\approx5.63\ sec$ (discard the zero answer)
