Question

The maximum height attained by a projectile when thrown at an angle $\theta$ with the horizontal is found to be half the horizontal range. Then $\theta$ is equal to?

Answer 1

$N=\frac{U^2{Sin\theta }^2}{2g}$

$R=\frac{U^2{Sin\theta }^2}{2g}$

$\frac{U^2{Sin\theta }^2}{2g}=\frac{1}{2}{U}^2\frac{2\theta sin\theta cos\theta }{g}$

$Sin\theta =2Cos\theta$

$Tan\theta =2$

$Q={Tan}^{-1}2$