Question

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projection is:

• A. $\theta ={\tan }^{-1}\left(\frac{1}{4}\right)$
• B. $\theta ={\tan }^{-1}\left(4\right)$
• C. $\theta ={\tan }^{-1}\left(2\right)$
• D. $\theta ={45}^o$

Answer 1

The correct option is B $\theta ={\tan }^{-1}\left(4\right)$

WE know that the maximum height $H=u^{2}Si{n}^2\theta /2g$ and horizontal range $R$ is given as $R=u^{2}Sin2\theta /g=2u^{2}Sin\theta Cos\theta /g$

Equating Both $H$ and $R$, we get $Si{n}^2\theta /2=2Sin\theta Cos\theta$ or $Tan\theta =4$ thus angle of projection $\theta =ta{n}^{-1}4$