Answer to Question #334511 in Physics for Alemayehu kumsa

Question

A projectile is fired in such away that its horizontal range is equal to its maximum height what is the angle of projection ?

Answer 1

$$H_{max}=L_{max}$$

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

$$\tan\theta=4$$

$$\theta=76^\circ$$