Answer to Question #160623 in Mechanics | Relativity for joel

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 expression for horizontal range of projectile is

$R=\frac{2v_0^2 sin(\theta)cos(\theta) }{g}$

The maximum height is

$H=\frac{(v_0sin(\theta))^2}{2g}$

Given that $R=3H$ . So

$\frac{2v_0^2 sin(\theta)cos(\theta) }{g}=3\frac{(v_0sin(\theta))^2}{2g}$

$tan\theta =\frac{4}{2}$

So the angle of projection is $\theta =tan^{-1}(\frac{4}{3})=53^o$