Question

If $R$ and $H$ represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is

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

Answer 1

The correct option is B

${\tan }^{-1}\left(\frac{2H}{R}\right)$

Explanation for the correct option:

Step 1: Apply the formula of projectile motion

In a projectile motion, the maximum height is given by $H=\frac{u^2{\sin }^2\theta }{2g}.......\left(i\right)$

and the horizontal range is $R=\frac{u^2\sin 2\theta }{2g}=\frac{2u^2\sin \theta \cos }{2g}.......\left(ii\right)$

Where, $u$ is the velocity of the projectile, which makes an angle $\theta$ with the horizontal component.

Step 2: Divide equation (i) by (ii), and we get

$\begin{array}{l}\Rightarrow \frac{H}{R}=\frac{\frac{u^2{\sin }^2\theta }{2g}}{\frac{2u^2\sin \theta \cos }{2g}}\\ \Rightarrow \frac{H}{R}=\frac{u^2{\sin }^2\theta }{2g}\times \frac{2g}{2u^2\sin \theta \cos }\\ \Rightarrow \frac{H}{R}=\frac{\sin \theta }{2\cos \theta }\\ \Rightarrow \frac{H}{R}=\frac{\tan \theta }{2}\\ \Rightarrow \tan \theta =\frac{2H}{R}\\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{2H}{R}\right)\end{array}$

Therefore, the angle of projectile is ${\tan }^{-1}\left(\frac{2H}{R}\right)$.

Hence, the correct option is (B).