Question

A projectile is fired in such a way that its horizontal range is equal to its maximum height. What is the angle of projection?

Answer 1

Equate them to figure out:

$$R=H,\\\space\\ \frac{v^2\sin(2\theta)}{g}=\frac{v^2\sin\theta}{2g},\\\space\\ 4\sin\theta\cos\theta=\sin\theta,\\\space\\ \theta=\arccos\frac 14=75.5°.$$