Answer
a)
The time after which the ball strikes the ground is $6\text{ seconds}$.
b) The ball reaches its maximum height of $\text{196 }\!\!~\!\!\text{ feet}$ in $\text{2}\text{.5 }\!\!~\!\!\text{ seconds}$.
Work Step by Step
(a)
We know that when the ball hits the ground, the displacement covered will be zero. So, equate
$\begin{align}
& s\left( t \right)=0 \\
& -16{{t}^{2}}+80t+96=0
\end{align}$
$\begin{align}
& -16{{t}^{2}}+80t+96=0 \\
& -{{t}^{2}}+5t+6=0 \\
& {{t}^{2}}-5t-6=0 \\
& {{t}^{2}}-6t+t-6=0
\end{align}$
$\begin{align}
& t\left( t-6 \right)+1\left( t-6 \right)=0 \\
& \left( t-6 \right)\left( t+1 \right)=0
\end{align}$
$ t=6\text{ or }t=-1$
Since t cannot be negative so, $ t=6\text{ }$
Thus, the ball strikes the ground after $6$ seconds.
(b)
We have to differentiate $ s\left( t \right)=-16{{t}^{2}}+80t+96$ to get the value of maximum height as:
$\begin{align}
& s\left( t \right)=-16{{t}^{2}}+80t+96 \\
& \frac{ds\left( t \right)}{dt}=-32t+80 \\
\end{align}$
And for maximum height, evaluate $\frac{ds\left( t \right)}{dt}=0$ as:
$\begin{align}
& -32t+80=0 \\
& 32t=80 \\
& t=\frac{80}{32} \\
& t=2.5
\end{align}$
Therefore, at $ t=2.5$, the value of s is:
$\begin{align}
& s\left( t \right)=-16{{\left( 2.5 \right)}^{2}}+80\left( 2.5 \right)+96 \\
& =-16\left( 6.25 \right)+200+96 \\
& =-100+296 \\
& =196
\end{align}$
Thus, the ball reaches its maximum height of $196$ feet at $2.5$ seconds.
