Answer
It reaches the ground after 5 s,
with downward speed 160 ft/s.
Work Step by Step
Solve for t when s(t)=0.
$400-16t^{2}=0\qquad/\div 16\\\\$
$25-t^{2}=0$
... Factor the LHS, difference of squares
$(5-t)(5+t)=0$
$t=5$ and
$t=-5$ (we discard the negative time)
In part (a) we found: $v(t)=-32t\quad[ft/s]$
$v(5)=-32\cdot 5=-160 $ ft/s.
(negative=downward)
