A batsman hits a ball with a velocity 'v', making an angle of 60°… | MHT CET 2026 (15 Apr, Shift 2)

Q1MHT CET 2026 · 15 APR · SHIFT 2Medium

Velocity, Momentum & Energy in Flight

A batsman hits a ball with a velocity 'v', making an angle of $60^\circ$ with the vertical. After some time direction of velocity is making an angle of $60^\circ$ with the horizontal. The speed of the ball at this instant is

$\left[\cos(60^\circ) = \frac{1}{2},\ \cos(30^\circ) = \frac{\sqrt{3}}{2}\right]$

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Answer: (b)

$60^\circ$ with the vertical means $30^\circ$ with the horizontal, so $v_x = v\cos 30^\circ = \frac{\sqrt{3}}{2}v$ .
The horizontal component stays constant throughout the flight.
When the velocity makes $60^\circ$ with the horizontal: $v'\cos 60^\circ = \frac{\sqrt{3}}{2}v$
$v' = \frac{\sqrt{3}v/2}{1/2} = \sqrt{3}v$
So the answer is (b) $\sqrt{3}v$ .

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Velocity, Momentum & Energy in Flight question from MHT CET 2026 (15 Apr, Shift 2)

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