Answer
a) $4.2$ feet
b) $0.744$ seconds and $3.006$ seconds
c) $60.45$ feet
d) $3.819$ seconds
Work Step by Step
Given \begin{equation}
h(t)=-16 t^2+60 t+4.2.
\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+60 \cdot(0)+4.2 =4.2.
\end{aligned}
\end{equation} The ball is hit at a height of $4.2$ ft.
b) Set $h(t)=40$ to find the time.
\begin{equation}
\begin{aligned}
&-16 t^2+60 t+4.2=40 \\
& -16 t^2+60 t+4.2-40=0 \\
& \frac{-16 t^2+60 t-35.8}{-16}=\frac{0}{-16} \\
& t^2-3.75 t+2.2375=0
\end{aligned}
\end{equation} Use the quadratic formula with $a= 1,\quad b= -3.75\ ,\quad c = 2.2375$.
\begin{equation}
\begin{aligned}
&t= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
& t=\frac{-(-3.5) \pm \sqrt{(-3.75)^2-4(1)(2.2375)}}{2(1)} \\
& =\frac{3.75 \pm \sqrt{5.1125}}{2}\\
&= \frac{3.75\pm 2.261}{2}\\
&= 1.875\pm 1.131\\
\end{aligned}
\end{equation} The solutions are: \begin{equation}
\begin{aligned}
& t=1.875-1.131=0.744 \\
& t=1.875+1.131=3.006.
\end{aligned}
\end{equation} The ball reached a height of $20$ feet after $0.744$ seconds and after $3.006$ seconds.
c) The time when the ball reached its maximum height is found from:
\begin{equation}
t=\frac{-b}{2 a}=\frac{-60}{2(-16)}=1.875.
\end{equation} The ball reached its maximum height aster $1.75$ seconds.
The maximum height of the ball is found from:
\begin{equation}
\begin{aligned}
h(1.875) & =-16 \cdot(1.875)^2+60 \cdot(1.875)+4.2 =60.45.
\end{aligned}
\end{equation} The maximum height of the ball is $60.45$ feet.
d) Set $h=0$ to find the time the ball hit the ground.
\begin{equation}
\begin{aligned}
&-16 t^2+60 t+4.2= 0\\
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
& t=\frac{-(60) \pm \sqrt{(60)^2-4(-16)(4.2)}}{2(-16)} \\
& =\frac{-60 \pm \sqrt{3868.8}}{-32}\\
&=-(-1.875 \pm 1.944 )
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
t & = 1.875+1.944 \\
& =3.819 \\
t & =1.875-1.944 \\
& =-0.069\\
\end{aligned}
\end{equation} The ball will hit the ground after about $3.819$ seconds later after being hit.
