Q129 / 180 · COMEDK 2021 · Physics
The angle of projection with the horizontal in terms of maximum height attained and horizontal range is given by
- A$\tan ^{-1}\left(\dfrac{2 H}{3 R}\right)$
- B$\tan ^{-1}\left(\dfrac{4 R}{3 H}\right)$
- Correct answerC$\tan ^{-1}\left(\dfrac{4 H}{R}\right)$
- D$\tan ^{-1}\left(\dfrac{R}{H}\right)$
Solution
The maximum height $H$ attained by a projectile is given by $H = \dfrac{u^2 \sin^2 \theta}{2g}$.
The horizontal range $R$ of the projectile is given by $R = \dfrac{u^2 \sin(2\theta)}{g} = \dfrac{2u^2 \sin \theta \cos \theta}{g}$.
Dividing the expression for $H$ by the expression for $R$, we get:
$\dfrac{H}{R} = \dfrac{\dfrac{u^2 \sin^2 \theta}{2g}}{\dfrac{2u^2 \sin \theta \cos \theta}{g}}$
$\dfrac{H}{R} = \dfrac{u^2 \sin^2 \theta}{2g} \times \dfrac{g}{2u^2 \sin \theta \cos \theta}$
$\dfrac{H}{R} = \dfrac{\sin \theta}{4 \cos \theta} = \dfrac{1}{4} \tan \theta$
Rearranging the terms to solve for $\tan \theta$:
$\tan \theta = \dfrac{4H}{R}$
Therefore, the angle of projection $\theta$ is given by:
$\theta = \tan^{-1}\left(\dfrac{4H}{R}\right)$
Answer: $\tan ^{-1}\left(\dfrac{4 H}{R}\right)$
The horizontal range $R$ of the projectile is given by $R = \dfrac{u^2 \sin(2\theta)}{g} = \dfrac{2u^2 \sin \theta \cos \theta}{g}$.
Dividing the expression for $H$ by the expression for $R$, we get:
$\dfrac{H}{R} = \dfrac{\dfrac{u^2 \sin^2 \theta}{2g}}{\dfrac{2u^2 \sin \theta \cos \theta}{g}}$
$\dfrac{H}{R} = \dfrac{u^2 \sin^2 \theta}{2g} \times \dfrac{g}{2u^2 \sin \theta \cos \theta}$
$\dfrac{H}{R} = \dfrac{\sin \theta}{4 \cos \theta} = \dfrac{1}{4} \tan \theta$
Rearranging the terms to solve for $\tan \theta$:
$\tan \theta = \dfrac{4H}{R}$
Therefore, the angle of projection $\theta$ is given by:
$\theta = \tan^{-1}\left(\dfrac{4H}{R}\right)$
Answer: $\tan ^{-1}\left(\dfrac{4 H}{R}\right)$
Hence, the correct answer is option: 'C'.