Q103 / 150 · MHT CET 2026 (17 April Shift 1) · Physics
Two balls A and B are projected at an angle of $45^\circ$ and $60^\circ$ respectively, so that the maximum heights reached are same for both. The ratio of initial velocity of projection of ball A to that for ball B is
$(\sin 30^\circ = \cos 60^\circ = \dfrac{1}{2},\ \sin 45^\circ = \cos 45^\circ = \dfrac{1}{\sqrt{2}},\ \sin 60^\circ = \cos 30^\circ = \dfrac{\sqrt{3}}{2})$
$(\sin 30^\circ = \cos 60^\circ = \dfrac{1}{2},\ \sin 45^\circ = \cos 45^\circ = \dfrac{1}{\sqrt{2}},\ \sin 60^\circ = \cos 30^\circ = \dfrac{\sqrt{3}}{2})$
- A$2 : \sqrt{3}$
- B$\sqrt{3} : 2$
- C$\sqrt{2} : \sqrt{3}$
- Correct answerD$\sqrt{3} : \sqrt{2}$
Solution
The maximum height reached by a projectile is given by $H = \dfrac{u^2 \sin^2 \theta}{2g}$
For ball A, $H_A = \dfrac{u_A^2 \sin^2 45^\circ}{2g}$
For ball B, $H_B = \dfrac{u_B^2 \sin^2 60^\circ}{2g}$
Since $H_A = H_B$, we have:
$\dfrac{u_A^2 \sin^2 45^\circ}{2g} = \dfrac{u_B^2 \sin^2 60^\circ}{2g}$
$u_A \sin 45^\circ = u_B \sin 60^\circ$
$\dfrac{u_A}{u_B} = \dfrac{\sin 60^\circ}{\sin 45^\circ}$
Substituting the values $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$ and $\sin 45^\circ = \dfrac{1}{\sqrt{2}}$:
$\dfrac{u_A}{u_B} = \dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{\sqrt{2}}} = \dfrac{\sqrt{3}}{\sqrt{2}}$
The ratio of initial velocity of projection of ball A to that for ball B is $\sqrt{3} : \sqrt{2}$.
For ball A, $H_A = \dfrac{u_A^2 \sin^2 45^\circ}{2g}$
For ball B, $H_B = \dfrac{u_B^2 \sin^2 60^\circ}{2g}$
Since $H_A = H_B$, we have:
$\dfrac{u_A^2 \sin^2 45^\circ}{2g} = \dfrac{u_B^2 \sin^2 60^\circ}{2g}$
$u_A \sin 45^\circ = u_B \sin 60^\circ$
$\dfrac{u_A}{u_B} = \dfrac{\sin 60^\circ}{\sin 45^\circ}$
Substituting the values $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$ and $\sin 45^\circ = \dfrac{1}{\sqrt{2}}$:
$\dfrac{u_A}{u_B} = \dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{\sqrt{2}}} = \dfrac{\sqrt{3}}{\sqrt{2}}$
The ratio of initial velocity of projection of ball A to that for ball B is $\sqrt{3} : \sqrt{2}$.
Hence, the correct answer is option: 'D'.