Question

If the horizontal range of a projectile is 4 times its maximum height which is the angle of projection

Answer 1

Horizontal range, $R= \frac{u^2sin2\theta}{g}$

maximum height, $H= \frac{u^2sin^2\theta}{2g}$

Given that R = 4H

$\Rightarrow \frac{u^2sin2\theta}{g} =4 \times \frac{u^2sin^2\theta}{2g} \\ \\ \Rightarrow \frac{2 u^2sin\theta cos \theta }{g} =4 \times \frac{u^2sin^2\theta}{2g}\\ \\ \Rightarrow \frac{2 u^2sin\theta cos \theta }{g} = \frac{2u^2sin^2\theta}{g}\\ \\ \Rightarrow cos \theta=sin\theta\\ \\ \Rightarrow \frac{sin\theta}{cos \theta}=tan \theta=1\\ \\ \Rightarrow \theta = 45^o$

Angle of projection is 45°