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

Answer 1

Answer: (A)

Maximum height, $H_0=\frac{u^2{\sin }^2\theta }{2g}$
Range, $R=\frac{u^2\sin 2\theta }{g}=\frac{2u^2\sin \theta \cos \theta }{g}$
Given, $H_0=\frac{R}{2}$
$\therefore \frac{u^2{\sin }^2\theta }{2g}=\frac{2u^2\sin \theta \cos \theta }{2g}$
$\Rightarrow \sin \theta =2\cos \theta$
$\Rightarrow \tan \theta =2$
$\therefore \theta ={\tan }^{-1}(2)$