Answer to Question #261570 in Mechanics | Relativity for tamrat

Question

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

Answer 1

$$R=3H$$ $$\frac{v^2\sin{2\theta}}{g}=2\frac{v^2\sin{\theta}\cos{\theta}}{g}=3 \frac{v^2\sin^2{\theta}}{2g}$$ $$\tan \theta=\frac{4}{3}$$ $$\theta=53\degree$$