Answer
(a) $2.75\ sec$
(b) $169\ ft$
(c) between 0.7 and 4.8 seconds.
(d) $6\ sec$
Work Step by Step
Given $s(t)=-16t^2+88t+48$,
(a) We have $a=-16, b=88$, the rocket will reach its maximum at $t=-\frac{b}{2a}=-\frac{88}{2(-16)}=\frac{11}{4}=2.75\ sec$
(b) At $t=2.75\ sec$, the maximum height is $s(2.75)=-16(2.75)^2+88(2.75)+48=169\ ft$
(c) Let $s(t)\gt 100$, we have $-16t^2+88t+48\gt 100$ or $-16t^2+88t-52\gt 0$. Graph the function $f(t)=-16t^2+88t-52$ as shown in the figure, we can find the solution set to the inequality as $(0.7,4.8)$ that is between 0.7 and 4.8 seconds.
(d) The rocket will hit the round when $s(t)=0$, we have $-16t^2+88t+48=0$ or $2t^2-11t-6=0$ which can be factored as $(x-6)(2x+1)=0$, thus $t=6\ sec$ (discard the negative answer).
