Answer
a) $4$ feet
b) $0.5$ seconds and $2$ seconds
c) $1.25$ seconds
d) $29$ feet
e) $2.596$ seconds
Work Step by Step
Given \begin{equation}
h(t)=-16 t^2+40 t+4.
\end{equation} a) Set $t=0$ to find the height of the ball when it is hit.
\begin{equation}
\begin{aligned}
h(0) & =-16 \cdot(0)^2+40 \cdot(0)+4 =4.
\end{aligned}
\end{equation} The ball is hit at a height of $4$ ft .
b) Set $h(t)=20$ to find the time.
\begin{equation}
\begin{aligned}
& -16 t^2+40 t+4=20 \\
& -16 t^2+40 t+4-20=0 \\
& \frac{-16 t^2+40 t-16}{-16}=\frac{0}{-16} \\
& t^2-2.5 t+1=0
\end{aligned}
\end{equation} Use the quadratic formula with $a= 1,\quad b= -2.5\ ,\quad c = 1$.
\begin{equation}
\begin{aligned}
&t= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
& t=\frac{-(-2.5) \pm \sqrt{(-2.5)^2-4(1)(1)}}{2(1)} \\
& =\frac{2.5 \pm \sqrt{2.25}}{2}\\
&= \frac{2.5\pm 1.5}{2}\\
&= 1.25\pm 1.5.
\end{aligned}
\end{equation} The solutions are \begin{equation}
\begin{aligned}
& t=1.25-0.75=0.5 \\
& t=1.25+0.75=2
\end{aligned}
\end{equation} The ball reached a height of $20$ feet after $0.5$ seconds and after $2$ seconds.
c) The time when the ball reached its maximum height is found from:
\begin{equation}
t=\frac{-b}{2 a}=\frac{-40}{2(-16)}=1.25.
\end{equation} The ball reached its maximum height aster $1.25$ seconds.
d) The maximum height of the ball is found from:
\begin{equation}
\begin{aligned}
h_{\text {max }} & =-16 \cdot(1.25)^2+40 \cdot(1.25)+4 =29\ \mathrm{ ft}.
\end{aligned}
\end{equation}
The maximum height of the ball is $29$ feet.
e) Set $h=0$ to find the time the ball hit the ground.
\begin{equation}
\begin{aligned}
-16 t^2+40 t+4&= 0\\
\frac{-16 t^2+40 t+4}{-16}&=\frac{0}{-16} \\
t^2-2.5 t-0.25&=0\\
& t=\frac{-(-2.5) \pm \sqrt{(-2.5)^2-4(1)(-0.25)}}{2(1)} \\
& =\frac{2.5 \pm \sqrt{7.25}}{2}\\
&=1.25 \pm 1.346
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
t & =1.25-1.346 \\
& =-0.096 \\
t & =1.25+1.346 \\
& =2.596
\end{aligned}
\end{equation} The ball will hit the ground after about $2.596$ seconds later after being hit.
