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:

• A. ${\tan }^{-1}\left(2\right)$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. ${\tan }^{-1}\left(\frac{1}{2}\right)$

Answer 1

The correct option is A ${\tan }^{-1}\left(2\right)$

The maximum height attained by a projectile = $H=\frac{u_y^2}{2g}$

The horizontal range of a projectile = $R=\frac{2u_x{u}_y}{g}$

It is given that $H=\frac{R}{2}$

$⟹\frac{u_y^2}{2g}=\frac{1}{2}\times \frac{2u_x{u}_y}{g}$

$⟹u_y=2u_x$

$⟹usin\theta =2ucos\theta$

$⟹tan\theta =2$

$⟹\theta =ta{n}^{-1}\left(2\right)$