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

$$R=3H$$

$$\frac{v^2\sin{2\theta}}{g}=2\frac{v^2\sin{\theta}\cos{\theta}}{g}=3 \frac{v^2\sin^2{\theta}}{2g}$$

$$\tan \theta=\frac{4}{3}$$

$$\theta=53\degree$$