Answer to Question #224441 in Physics for Ladepa63

Question

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

Answer 1

Given:

$R=3H$

The range

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

the maximum height

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

Hence,

$$\frac{v_0^2\sin2\theta}{g}=3*\frac{v_0^2\sin^2\theta}{2g}$$

$$\sin2\theta=3/2\sin^2\theta$$

$$2\sin\theta\cos\theta=3/2\sin^2\theta$$

$$\tan\theta=4/3$$

$$\theta=53^{\circ}$$