Answer
(a) The ball would reach half of the distance to the ground ($48feet$) in about $1.73sec$
(b) The ball reaches ground in about $2.45$ seconds.
Work Step by Step
According to the problem, we have a formula: $h=-16t^2+h_{0}$
Where $h_{0} $ represents height the object starts falling from and $h$ is height of the object to given $t$ time.
Our object starts falling from $96ft$, so we have: $h=-16t^2+96$
(a) Half of the distance to ground is $\frac{96}{2}=48feet$, which means that $h=48$:
$48=-16t^2+96$
$16t^2=48$
$t^2=3$
$t=\sqrt{3}\approx1.73sec$
(b) We have to measure time it takes to reach ground level, that means $h=0$, so we get the following expression:
$0=-16t^2+96$
$16t^2=96$
$t^2=6$
$t=\sqrt{6}\approx2.45 sec$
