Answer
a) $96$
b) The solutions are $t \approx 1.382$ and $t \approx 3.618$.
See the graph.
Work Step by Step
(a) $h(2)=80(2)-16(2)^2=160-64=96$. This means that after 2 seconds, the ball's height is 96 feet.
b) Set $h(t) = 80$ and solve for $t$:
$$
\begin{aligned}
80 t-16 t^2 & =80 \\
16 t^2-80 t+80 & =0\\
t^2-5 t+5&=0\\
t&=\frac{5 \pm \sqrt{25-4 \cdot 5}}{2}\\
&=\frac{5 \pm \sqrt{5}}{2}
\end{aligned}
$$
The solutions are $t \approx 1.382$ and $t \approx 3.618$. This means that the ball reaches the height of 80 ft once on the way up, after about 1.382 seconds, and once on the way down, after about 3.618 seconds.
See the graph.
