Answer
t=2.56 seconds
|v|=65.97 feet
Work Step by Step
It is given that, $v_0=16$ and $s(0)=64$.
We know that,
$s(t)=v_0t + \frac{1}{2}at^2+s(0)$
Putting in the values,
$s(t)=16t -16t^2+64$
When it hits the ground, $s(t_0) = 0.$
$16t_0 -16t_0^2+64=0 $
or $-t_0^2+t_0+4=0$
or $t_0= \frac{1+\sqrt 17}{2} = 2.56$, discarding the negative value of t.
$v(t) = \frac{ds(t)}{dt}=16 - 32t$
therefore,$ v(t_0) =16-32( \frac{1+\sqrt 17}{2}) = -65.97$
