Answer
$16,000 \ \ \mathrm{f}\mathrm{t}/\mathrm{s}$
Work Step by Step
By exercise 70, $s(t)=-16t^{2}+v_{0}t+h_{0}.$
We take the initial height to be 0, $\quad h_{0}=0.$
so
$s(t)=-16t^{2}+16,000t.$
We want the t when the projectile hits the ground, $s(t)=0$
$-16t^{2}+16,000t=0$
$-16t(t-1000)=0$
$t=0$ or $1000$
We take $t=1000$, because $t=0$ is the initial takeoff time.
The velocity is given with $v(t)=-32t+v_{0}$
$v(t)=-32t+16,000$
At $t=1000$,
$v(1000)=-32(1000)+16,000$
$=-16,000\ \ \mathrm{f}\mathrm{t}/\mathrm{s}$ .
(the minus indicates downward direction)
The speed is
$16,000 \ \ \mathrm{f}\mathrm{t}/\mathrm{s}$
