Answer:
2 seconds
Step-by-step explanation:
Filling in the given values in the appropriate ballistic motion formula, we have ...
[tex]h(t)=-16t^2+v_0t+s_0\qquad\text{initial vertical velocity $v_0$, and initial height $s_0$}\\\\h(t)=-16t^2+64\qquad\text{$v_0=0$ and $s_0=64$}\\\\t^2=4\qquad\text{divide by 16, add $t^2$}\\\\t=2\qquad\text{take the positive square root}[/tex]
It will take the ball 2 seconds to hit the ground.