Answer
$158\text{ ft}$
Work Step by Step
The bridge is in the form of a parabola, where $x=0$, and the bridge is $160\text{ ft}$; then, the bridge is from $-80\text{ to }80$.
The vertical cables are $30\text{ ft}$above road level at the midpoint of the bridge, so, the point would be $\left( 0,30 \right)$ on the $y-axis$.
Also, where the vertical cables are $80\text{ ft}$ above road level at a point 50ft from the midpoint, the point would be $\left( 50,80 \right)\,\text{ and}\,\,\left( -50,80 \right)$
As it is known that the standard form of the equation is $a{{x}^{2}}+bx+c=0$
Now, forming the quadratic equation that fits the points $\left( 0,30 \right)$,$\left( 50,80 \right)\,\text{ and}\,\,\left( -50,80 \right)$:
$\begin{align}
& a{{\left( 0 \right)}^{2}}+b\left( 0 \right)+c=30 \\
& 0+0+c=30 \\
& c=30
\end{align}$
$\begin{align}
& a{{\left( 50 \right)}^{2}}+b\left( 50 \right)+c=80 \\
& 2500a+50b+c=80
\end{align}$
$2500a+50b+c=80$ …… (1)
$\begin{align}
& a{{\left( -50 \right)}^{2}}+b\left( -50 \right)+c=80 \\
& 2500a-50b+c=80
\end{align}$
$2500a-50b+c=80$ …… (2)
Solve the above equation for a, b and c by substituting the value of c in equation (1) and (2).
$\begin{align}
& 2500a+50b+30=80 \\
& 2500a+50b=50 \\
& 50a+b=1
\end{align}$
$\begin{align}
& 2500a-50b+30=80 \\
& 2500-50b=50 \\
& 50a-b=1
\end{align}$
So, upon solving the equations, the value are $a=\frac{1}{50}$and $b=0$,
So the equation of the bridge would be:
$\begin{align}
& y=a{{x}^{2}}+bx+c \\
& y=\left( \frac{1}{50} \right){{x}^{2}}+\left( 0 \right)b+30 \\
& y=\left( \frac{1}{50} \right){{x}^{2}}+30
\end{align}$
In order to find the longest vertical cable, the furthest points will be taken, which are $-80\text{ and }80$
So, put $x=80$ in the equation $y=\frac{1}{50}{{x}^{2}}+30$.
$\begin{align}
& y=\frac{1}{50}{{\left( 80 \right)}^{2}}+30 \\
& y=158\text{ ft} \\
\end{align}$
Thus, the length of the longest vertical cable is $158\text{ ft}$.
