Q110 / 150 · MHT CET 2026 (20 April Shift 2) · Physics
Two spheres are projected at angles $30^\circ$ and $45^\circ$ with the horizontal. The maximum height reached by both is same. The ratio of their initial velocities is, $\left(\sin 45^\circ = \dfrac{1}{\sqrt{2}}, \sin 30^\circ = 0.5\right)$
- A$2 : 3$
- Correct answerB$\sqrt{2} : 1$
- C$3 : 1$
- D$\sqrt{2} : \sqrt{3}$
Solution
The maximum height reached by a projectile is given by $H = \dfrac{u^2 \sin^2 \theta}{2g}$
Given that the maximum heights are equal for both spheres:
$H_1 = H_2$
$\dfrac{u_1^2 \sin^2 \theta_1}{2g} = \dfrac{u_2^2 \sin^2 \theta_2}{2g}$
$u_1 \sin \theta_1 = u_2 \sin \theta_2$
Substituting $\theta_1 = 30^\circ$ and $\theta_2 = 45^\circ$:
$u_1 \sin 30^\circ = u_2 \sin 45^\circ$
$u_1 \left(\dfrac{1}{2}\right) = u_2 \left(\dfrac{1}{\sqrt{2}}\right)$
$\dfrac{u_1}{u_2} = \dfrac{2}{\sqrt{2}} = \dfrac{\sqrt{2}}{1}$
The ratio of their initial velocities is $\sqrt{2} : 1$.
Answer: $\sqrt{2} : 1$
Given that the maximum heights are equal for both spheres:
$H_1 = H_2$
$\dfrac{u_1^2 \sin^2 \theta_1}{2g} = \dfrac{u_2^2 \sin^2 \theta_2}{2g}$
$u_1 \sin \theta_1 = u_2 \sin \theta_2$
Substituting $\theta_1 = 30^\circ$ and $\theta_2 = 45^\circ$:
$u_1 \sin 30^\circ = u_2 \sin 45^\circ$
$u_1 \left(\dfrac{1}{2}\right) = u_2 \left(\dfrac{1}{\sqrt{2}}\right)$
$\dfrac{u_1}{u_2} = \dfrac{2}{\sqrt{2}} = \dfrac{\sqrt{2}}{1}$
The ratio of their initial velocities is $\sqrt{2} : 1$.
Answer: $\sqrt{2} : 1$
Hence, the correct answer is option: 'B'.