91Ó°ÊÓ

The given equation of the distance of the ball from the ground is$s\left(t\right)=96t-16t^2$

• Substitute 0 for $s\left(t\right)$into the given function and then factorize.

role="math" localid="1646425677585" $\begin{array}{rcl}s\left(t\right)& =& 96t-16t^2\\ & =& 16t(6-t)\end{array}$

• Equate both the factors with 0.

role="math" localid="1646425671158" $\begin{array}{rcl}16t& =& 0\\ t& =& 0\\ 6-t& =& 0\\ t& =& 6\end{array}$

So, the ball hits the ground after 6 seconds.

• The given equation is $s\left(t\right)=96t-16t^2$.
• The minimum height of the ball is 128 feet.

• The minimum height of the ball is 128 feet. So, $96t-16t^2>128\mathrm{or}-16t^2+96t-128>0$.
• Find the factors of the function.

$\begin{array}{rcl}-16t^2+96t-128& =& 0\\ -16(t-2)(t-4)& =& 0\end{array}$

• Equate the factors to 0.

$\begin{array}{rcl}t-2& =& 0\\ t& =& 2\\ t-4& =& 0\\ t& =& 4\end{array}$

• So, the 0- intercepts are $(2,0),(4,0)$.
• The value of $f\left(0\right)=-128$.
• So, the y- intercept is $(0,-128)$.
• The vertex of the parabola is at $x=\frac{-b}{2a}=\frac{-96}{-32}=3$.
• The value of $f\left(3\right)=16$.
• So, the vertex of the parabola is at $(3,16)$.

• Use the intercepts and the vertex to plot the graph of the function.

• From the graph, the curve is above the horizontal axis when $2<t<4$.
• So, the solution set is$\left\{t|2<t<4\right\}$