Answer
$\frac{\pi}{4}$
Work Step by Step
Step 1. We are given the range formula $R=\frac{v_0^2}{g}sin2\alpha$ and we need to find the critical $\alpha$ which gives a maximum range, where $0\lt \alpha\lt \frac{\pi}{2}$.
Step 2. Take the derivative with respect to $\alpha$ to get $R'=\frac{2v_0^2}{g}cos2\alpha$ and let $R'=0$ to get $2\alpha=\frac{\pi}{2},\frac{3\pi}{2}$ and $\alpha=\frac{\pi}{4},\frac{3\pi}{4}$. Choose the answer in the domain to get $\alpha=\frac{\pi}{4}$ which gives $R=\frac{v_0^2}{g}$
Step 3. Check $R''(\frac{\pi}{4})=-\frac{4v_0^2}{g}\lt0$ indicating a concave down region with a local maximum.
