Question

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

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

Answer 1

Answer: (B)

Horizontal range, $R=\frac{u^2\sin 2\theta }{g}$
where $u$ is the velocity of projection and $\theta$ is the angle of projection
Maximum height, $H=\frac{u^2{\sin }^2\theta }{2g}$
According to question $R=H$
$\therefore \frac{u^2\sin 2\theta }{g}=\frac{u^2{\sin }^2\theta }{2g}$
$\frac{2u^2\sin \theta \cos \theta }{g}=\frac{u^2{\sin }^2\theta }{2g}$
$\tan \theta =4$
or $\theta ={\tan }^{-1}(4)$