Answer
(a) $1$ seconds, $5$ seconds
(b) $6$ seconds
Work Step by Step
Remember that the formula $s=-16t^2+v_{0}t$ describes the height, $s$ of an object thrown straight up at a velocity, $v_{0}$, after $t$ seconds.
With a given velocity $v_{0}$ of $96$ and the height $s$ of $80$ we can find how many seconds, $t$ the object took to get there.
$80=-16t^2+96t$
Rewrite in standard quadratic equation form: $ax^2 + bx +c=0$
$-16t^2+96t-80=0$
where $a=-16$, $b=96$, and $c=-80$
apply the quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
$x=\dfrac{(-)(96)\pm\sqrt{(96)^2-4(-16)(-80)}}{2(-16)}$
$x=\dfrac{-96\pm\sqrt{9216-5120}}{-32}$
$x=\dfrac{-96\pm\sqrt{4096}}{-32}$
$x=\dfrac{-96\pm64}{-32}$
$x=\dfrac{-96+64}{-32}$ or $x=\dfrac{-96-64}{-32}$
$x=\dfrac{-32}{-32}$ or $x=\dfrac{-160}{-32}$
$x=1$ or $x=5$
So after $1$ second the object reaches $80$ft, and after $5$ seconds it reaches $80$ft again on its way back down to the ground.
To find how long it takes the object to hit the ground, a height $s$ of $0$, we repeat the process with the new value
$0=-16t^2+96t$
$-16t^2+96t+0=0$
where $a=-16$, $b=96$, and $c=0$
apply the quadratic formula
$x=\dfrac{(-)(-96)\pm\sqrt{(96)^2-4(-16)(0)}}{2(-16)}$
$x=\dfrac{96\pm\sqrt{(96)^2-0}}{2(-16)}$
$x=\dfrac{96\pm96}{-32}$
$x=\dfrac{96+96}{-32}$ or $x=\dfrac{96-96}{-32}$
$x=\dfrac{192}{-32}$ or $x=\dfrac{0}{-32}$
$x=6$ or $x=0$
The objects starts at the ground at $0$ seconds, and hits the ground at $6$ seconds