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 $\theta$ =

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

Answer 1

Answer: (D)

$tan^{-1} (2)$

Answer 2

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