Answer to Question #104838 in Physics for Galata

Question

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

Answer 1

The horizontal range

$$R=\frac{v_0^2\sin2\theta}{g}.$$

The maximum height

$$H=\frac{v_0^2\sin^2\theta}{2g}.$$

Hence

$$\frac{R}{H}=\frac{2\sin2\theta}{\sin^2\theta}.$$ $$\cot\theta=\frac{R}{4H}=\frac{3}{4}.$$ $$\theta=53.1^{\circ}$$