Answer
$209.1$ meters dropped.
Work Step by Step
Before we try to solve the integral, we first need to find the boundaries of $x$, which is the instant the prey is dropped, and the instant the prey falls on the ground
$y=180-\frac{x^{2}}{45}$
$\frac{dy}{dx}=\frac{-2x}{45}$
When the prey is dropped at that instant, $x=0$, so $y=180$. To find the lower bound, we set $y$ to zero and solve for $x$.
$180-\frac{x^{2}}{45}=0$
$180=\frac{x^{2}}{45}$
$8100=x^{2}$
$x=90$
We now solve for the arc length
$A=\int_{0}^{90}\sqrt (1+ (\frac{-2x}{45})^2)dx$
$=\int_{0}^{90}\sqrt (1+ (\frac{-4x^{2}}{2025})^2)dx$
$\approx209.1m$ fallen.
