Answer
(a) the object never reaches a height of $80$ft
(b) $2$ 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 $32$ and the height $s$ of $80$ we can find how many seconds, $t$ the object took to get there.
$80=-16t^2+32t$
Rewrite in standard quadratic equation form: $ax^2 + bx +c=0$
$-16t^2+32t-80=0$
where $a=-16$, $b=32$, and $c=-80$
apply the quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
$x=\dfrac{(-)(32)\pm\sqrt{(32)^2-4(-16)(-80)}}{2(-16)}$
$x=\dfrac{-32\pm\sqrt{1064-5120}}{-32}$
$x=\dfrac{-32\pm\sqrt{-4056}}{-32}$
Since the discrimant, $-4056$ is negative we know that the solutions to this quadratic equation are non-real complex, therefore
The object never reaches a height of $80$ft
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
$0=-16t^2+32t$
$-16t^2+32t+0=0$
where $a=-16$, $b=32$, and $c=0$
apply the quadratic formula
$x=\dfrac{(-)(32)\pm\sqrt{(32)^2-4(-16)(0)}}{2(-16)}$
$x=\dfrac{-32\pm\sqrt{(32)^2-0}}{2(-16)}$
$x=\dfrac{-32\pm32}{-32}$
$x=\dfrac{-32+32}{-32}$ or $x=\dfrac{-32-32}{-32}$
$x=\dfrac{0}{-32}$ or $x=\dfrac{-64}{-32}$
$x=$ or $x=2$
The objects starts at the ground at $0$ seconds, and hits the ground at $2$ seconds
