Answer
a) $630$ feet
b) $630$ feet
Work Step by Step
a) We use the $x$-coordinates of the points where the arch meets the ground to find the width of the arch at ground level. At ground level, $y=0$. So we substitute $0$ for $y$ and solve for $x$:
$$\begin{align*}
y&=-\frac{2}{315}(x+315)(x-315)&&\text{Write equation.}\\
0&=-\frac{2}{315}(x+315)(x-315)&&\text{Substitute }0\text{ for }y\\
0&=(x+315)(x-315)&&\text{Multiply each side by }-\frac{315}{2}\\
x+315&=0\text{ or }x-315=0&&\text{Zero-Product Property.}\\
x&=-315\text{ or }x=315&&\text{Solve for }x.
\end{align*}$$ The width of the arch at ground level is the distance between the $x$-coordinates, $-315$ and $315$. So the width of the arch at ground level is $|-315-315|=630$ feet.
b) The height of the arch is the value of $y$ when $x=0$:
$$\begin{align*}
y&=-\frac{2}{315}(0+315)(0-315)\\
&=-\frac{2}{315}(315)(-315)\\
&=630.
\end{align*}$$ The arch is $630$ feet tall.
