Q103 / 150 · MHT CET 2026 (18 April Shift 2) · Physics
For a projectile motion, the range R is 'n' times the maximum height H. So the angle of projection is
- A$\sin^{-1}\left(\dfrac{2}{n}\right)$
- B$\cos^{-1}\left(\dfrac{4}{n}\right)$
- Correct answerC$\tan^{-1}\left(\dfrac{4}{n}\right)$
- D$\tan^{-1}\left(\dfrac{2}{n}\right)$
Solution
The horizontal range $R$ of a projectile is given by
$R = \dfrac{u^2 \sin 2\theta}{g} = \dfrac{2u^2 \sin\theta \cos\theta}{g}$
The maximum height $H$ is given by
$H = \dfrac{u^2 \sin^2\theta}{2g}$
Given that $R = nH$, substituting the expressions for $R$ and $H$
$\dfrac{2u^2 \sin\theta \cos\theta}{g} = n \left( \dfrac{u^2 \sin^2\theta}{2g} \right)$
Cancelling $\dfrac{u^2 \sin\theta}{g}$ from both sides
$2 \cos\theta = \dfrac{n \sin\theta}{2}$
$\dfrac{\sin\theta}{\cos\theta} = \dfrac{4}{n}$
$\tan\theta = \dfrac{4}{n}$
$\theta = \tan^{-1}\left(\dfrac{4}{n}\right)$
Answer: $\tan^{-1}\left(\dfrac{4}{n}\right)$
$R = \dfrac{u^2 \sin 2\theta}{g} = \dfrac{2u^2 \sin\theta \cos\theta}{g}$
The maximum height $H$ is given by
$H = \dfrac{u^2 \sin^2\theta}{2g}$
Given that $R = nH$, substituting the expressions for $R$ and $H$
$\dfrac{2u^2 \sin\theta \cos\theta}{g} = n \left( \dfrac{u^2 \sin^2\theta}{2g} \right)$
Cancelling $\dfrac{u^2 \sin\theta}{g}$ from both sides
$2 \cos\theta = \dfrac{n \sin\theta}{2}$
$\dfrac{\sin\theta}{\cos\theta} = \dfrac{4}{n}$
$\tan\theta = \dfrac{4}{n}$
$\theta = \tan^{-1}\left(\dfrac{4}{n}\right)$
Answer: $\tan^{-1}\left(\dfrac{4}{n}\right)$
Hence, the correct answer is option: 'C'.