Answer
a) $4$ feet
b) $0.427$ seconds and $3.073$ seconds
c) $1.75$ seconds
d) $53$ feet
Work Step by Step
Given \begin{equation}
h(t)=-16 t^2+56 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+56 \cdot(0)+4 =4.
\end{aligned}
\end{equation} The ball is hit at a height of $4$ ft .
b) Set $h(t)=25$ to find the time.
\begin{equation}
\begin{aligned}
& -16 t^2+56 t+4=25 \\
& -16 t^2+56 t+4-25=0 \\
& \frac{-16 t^2+56 t-21}{-16}=\frac{0}{-16} \\
& t^2-3.5 t+1.3125=0
\end{aligned}
\end{equation} Use the quadratic formula with $a= 1,\quad b= -3.5\ ,\quad c = 1.3125$.
\begin{equation}
\begin{aligned}
&t= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
& t=\frac{-(-3.5) \pm \sqrt{(-3.5)^2-4(1)(1.3125)}}{2(1)} \\
& =\frac{3.5 \pm \sqrt{7}}{2}\\
&= \frac{3.5\pm 2.646}{2}\\
&= 1.75\pm 1.323\\
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
& t=1.75-1.323=0.427 \\
& t=1.75+1.323=3.073
\end{aligned}
\end{equation} The ball reached a height of 20 feet after $0.427$ seconds and after $3.073$ seconds.
c)The time when the ball reached its maximum height is found from:
\begin{equation}
t=\frac{-b}{2 a}=\frac{-56}{2(-16)}=1.75
\end{equation} The ball reached its maximum height aster $1.75$ seconds.
d) The maximum height of the ball is found from:
\begin{equation}
\begin{aligned}
h_{\text {max }} & =-16 \cdot(1.75)^2+56 \cdot(1.75)+4 =53\ \mathrm{ft}.
\end{aligned}
\end{equation} The maximum height of the ball is $53$ feet.
