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, $H_{0}=\frac{u^{2} \sin ^{2} \theta}{2 g}$
Range, $R=\frac{u^{2} \sin 2 \theta}{g}=\frac{2 u^{2} \sin \theta \cos \theta}{g}$
Given, $H_{0}=\frac{R}{2}$
$\therefore \frac{u^{2} \sin ^{2} \theta}{2 g}=\frac{2 u^{2} \sin \theta \cos \theta}{2 g}$
$\Rightarrow \sin \theta=2 \cos \theta$
$\Rightarrow \tan \theta=2$
$\therefore \theta=\tan ^{-1}(2)$