Question

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

Answer 1

The range:

$$R=\frac{v^2\sin(2\theta)}{g}.$$

Height:

$$H=\frac{v^2\sin^2(\theta)}{2g}.$$

We know that

$$R=3H,\\\space\\ \frac{v^2\sin(2\theta)}{g}=3\frac{v^2\sin^2(\theta)}{2g},\\\space\\ 2\sin\theta\cos\theta=\frac32\sin^2\theta,\\\space\\ \tan\theta=\frac43,\\\space\\ \theta=53°.$$