Answer
$5$ feet
Work Step by Step
Given \begin{equation}
T(L)=2 \pi \sqrt{\frac{L}{32}}.
\end{equation} Set $T(L)= 2.5$ and solve for $L$.
\begin{equation}
\begin{aligned}
T(2)&=2.5\\
2 \pi \sqrt{\frac{L}{32}}& =2.5\\
\sqrt{\frac{L}{32}}&=\frac{2.5}{2\pi}\\
\left( \sqrt{\frac{L}{32}} \right)^{2}&=\left( \frac{1.25}{\pi} \right)^{2}\\
\frac{L}{32}&= \frac{1.5625}{\pi^2}\\
L&= \frac{32\cdot 1.5625}{\pi^2}\\
&= \frac{50}{\pi^2}\\
&\approx 5.
\end{aligned}
\end{equation} The length of the foot pendulum with period of $2.5$ seconds is about $5$ feet long.
