Since we are looking for length L, we rearrange the equation using L on the left side.
[tex]\begin{gathered} T=2\pi\sqrt[]{\frac{L}{g}} \\ 2\pi\sqrt[]{\frac{L}{g}}=T \\ \frac{2\pi\sqrt[]{\frac{L}{g}}}{2\pi}=\frac{T}{2\pi} \\ \mleft(\sqrt[]{\frac{L}{g}}=\frac{T}{2\pi}\mright)^2 \\ \frac{L}{g}=\frac{T^2}{4\pi^2} \\ L=\frac{T^2g}{4\pi^2} \end{gathered}[/tex]
Now that we have the equation for L, substitute using the following given:
g = 32 ft/s^2, and T = 1 second
[tex]\begin{gathered} L=\frac{T^2g}{4\pi^2} \\ L=\frac{(1s)^2(32\text{ ft/s}^2)}{4\pi^2},\text{ the }s^2\text{ cancels out} \\ L=\frac{32\text{ ft}}{4\pi^2} \\ L=\frac{8\text{ ft}}{\pi^2} \\ L=0.8105708384579\text{ ft} \\ \text{Convert to inches} \\ 0.8105708384579\text{ ft}\times\frac{12\text{ inch}}{1\text{ ft}} \\ =9.72685\text{ inch} \\ =10\text{ inch (rounded off to the nearest inch)} \end{gathered}[/tex]
