Answer
(a) $600\ ft$
(b) $25\ sec$.
(c) $10600\ ft$
(d) $(6.5,43.5)\ sec$
(e) $50.7\ sec$
Work Step by Step
(a) At $t=0$, we have $s(0)=600\ ft$ and this is the initial height.
(b) The maximum happens when $t=-\frac{800}{2(-16)}=25\ sec$.
(c) At the maximum, we have $s(25)=-16(25)^2+800(25)+600=10600\ ft$
(d) Let $-16(t)^2+800(t)+600\gt 5000$, graph $f(t)=-16(t)^2+800(t)-4500$ to find $t\in(6.5,43.5)\ sec$ for $f(t)\gt0$
(e) To hit the ground, we have $s(t)=0$, graph $s(t)$ to find $t=50.7\ sec$
