MHTCET 17 Apr 2026 Shift 1 Paper

Concept:For equal maximum heights, the initial velocity of a projectile is inversely proportional to the sine of its angle of projection.Explanation:The maximum height reached by a projectile is given by$$H = \frac{u^2 \sin^2 \theta}{2g}$$where $u$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.Since both balls reach the same maximum height, equate their height expressions:$$\frac{u_A^2 \sin^2 45^\circ}{2g} = \frac{u_B^2 \sin^2 60^\circ}{2g}$$Cancelling $2g$ from both sides,$$u_A^2 \sin^2 45^\circ = u_B^2 \sin^2 60^\circ$$Taking the positive square root on both sides,$$u_A \sin 45^\circ = u_B \sin 60^\circ$$Using $\sin 45^\circ = \frac{1}{\sqrt{2}}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$,$$\frac{u_A}{u_B} = \frac{\sin 60^\circ}{\sin 45^\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}} = \frac{\sqrt{3}}{\sqrt{2}}$$Therefore,$$u_A : u_B = \sqrt{3} : \sqrt{2}$$Answer:The required ratio of initial velocity of ball A to ball B is $\sqrt{3} : \sqrt{2}$.Hence, the correct option is D.