Answer
$h\left( d \right)=\left( -0.0068 \right){{d}^{2}}+\left( 0.8571 \right)d$
Work Step by Step
Let the arrow be launched from the point $\left( 0,0 \right)$.
Then, the point at the bottom of the tower will be $\left( 63,0 \right)$ and at the top of the tower will be
$\left( 63,27 \right)$.
In this case, if the arrow is launched at the point $(63,0)$, then the arrow will land at $\left( 126,0 \right)$.
Now, the points that emerged are $\left( 0,0 \right)$,$\left( 63,27 \right)$and $\left( 126,0 \right)$.
The standard form of the equation is:
$h\left( d \right)=a{{d}^{2}}+bd+c$, where $h$ is the height of the arrow and $d$ be the horizontal distance traveled.
So, use the three coordinates $\left( 0,0 \right)$,$\left( 63,27 \right)$ and $\left( 126,0 \right)$.
When $x=0,y=0$
$\begin{align}
& a{{\left( 0 \right)}^{2}}+b\left( 0 \right)+c=0 \\
& c=0
\end{align}$
The equation simplifies to:
$\begin{align}
& h\left( d \right)=a{{d}^{2}}+bd+0 \\
& h\left( d \right)=a{{d}^{2}}+bd \\
\end{align}$
When $x=63\text{ and }y=27$,
$\begin{align}
& a{{\left( 63 \right)}^{2}}+b\left( 63 \right)=27 \\
& 441a+7b=3
\end{align}$
$441a+7b=3$ …… (1)
When $x=126\text{ and }y=0$, then,
$\begin{align}
& a{{\left( 126 \right)}^{2}}+b\left( 126 \right)=0 \\
& 126\left( 126a+b \right)=0 \\
& 126a+b=0
\end{align}$
$126a+b=0$ …… (2)
Substitute the value of $b=-126a$ from equation (2) into equation (1),
$\begin{align}
& 441a+7\left( -126a \right)=3 \\
& 441a-882a=3 \\
& a\left( 441-882 \right)=3 \\
& -441a=3
\end{align}$
Divide by $-441$ on both sides,
$\begin{align}
& \frac{-441d}{-441}=\frac{3}{-441} \\
& d=\frac{-3}{441} \\
& =-0.0068
\end{align}$
Now, substitute the value of $a$ in equation (2):
$\begin{align}
& b=-126\left( -0.0068 \right) \\
& =0.8571
\end{align}$
Thus, the quadratic function is $h\left( d \right)=\left( -0.0068 \right){{d}^{2}}+\left( 0.8571 \right)d$.
