Answer
$a.\quad 6$ seconds
$b.\quad 5$ seconds
Work Step by Step
$a.$
On the ground, s=0.
$96+80t-16t^{2}=0$
Divide with -16 and write in standard form
$-6-5t+t^{2}=0$
$t^{2}-5t-6=0$
Factoring, $-6$ and $+1$ are factors of -6 and their sum is -5,
$(t+1)(t-6) =0$
Ignoring the negative solution,
$t=6 $ seconds.
$b.$
We want to find t when s=96
$96+80t-16t^{2}=96$
$80t-16t^{2}=0$
$16t(5-t)=0$
$t=0$ is the initial time (the ball was initially thrown up...)
$t=5$ s (is the time it takes for the ball to fall back down to the height from which it was thrown up)
