Answer
a) $t=\sqrt3$
b) $t=\sqrt6$
Work Step by Step
Using the given formula (and the fact that $t$ cannot be negative), we obtain:
a) We plug in half the distance of $h_o=96$ and solve for $t$:
$96/2=-16t^2+96\\-48=-16t^2\\t^2=3\\t=\sqrt3$
b) Now, we plug in the ground height $h=0$:
$0=-16t^2+96\\-96=-16t^2\\t^2=6\\t=\sqrt6$
