MHTCET 15 Apr 2026 Shift 2 Paper

Concept:In projectile motion, the horizontal component of velocity remains constant because there is no horizontal acceleration.Explanation:The ball is hit with speed $v$ making an angle of $60^{\circ}$ with the vertical.Therefore, its angle with the horizontal is:$$90^{\circ} - 60^{\circ} = 30^{\circ}$$The initial horizontal component of velocity is:$$u_x = v \cos 30^{\circ}$$Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$, we get:$$u_x = \frac{\sqrt{3}}{2} v$$After some time, the velocity makes an angle of $60^{\circ}$ with the horizontal.Let the speed at this instant be $V$.Its horizontal component then is:$$V \cos 60^{\circ}$$Because horizontal velocity is conserved:$$V \cos 60^{\circ} = \frac{\sqrt{3}}{2} v$$Using $\cos 60^{\circ} = \frac{1}{2}$:$$V \times \frac{1}{2} = \frac{\sqrt{3}}{2} v$$Solving for $V$:$$V = \sqrt{3} v$$Thus, the speed of the ball at that instant is $\sqrt{3} v$.This matches option B.Answer:$\sqrt{3} v$ (Option B)