Question

A projectile is fired in such a way that its horizontal range is equal to 3 times its maximum height. What is the angle of projection?

Answer 1

Write the equation that ties the velocity and angle of projection with the range:

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

The height:

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

We read that

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