A batsman hits a ball with a velocity 'v',mak | vedclass

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(B) In projectile motion, the horizontal component of velocity remains constant throughout the flight.Initially, the velocity $v$ makes an angle of $6

Vedclass · Accepted Answer

$\sqrt{3}v$

Vedclass · Answer

(B) In projectile motion, the horizontal component of velocity remains constant throughout the flight.Initially, the velocity $v$ makes an angle of $60^{\circ}$ with the vertical, which means it makes an angle of $90^{\circ} - 60^{\circ} = 30^{\circ}$ with the horizontal.Therefore, the horizontal component of velocity is $v_x = v \cos(30^{\circ}) = v \frac{\sqrt{3}}{2}$.At the later instant, the velocity $v'$ makes an angle of $60^{\circ}$ with the horizontal.The horizontal component of this new velocity is $v'_x = v' \cos(60^{\circ}) = v' \frac{1}{2}$.Since the horizontal component remains constant, $v_x = v'_x$.$v \frac{\sqrt{3}}{2} = v' \frac{1}{2}$.Solving for $v'$, we get $v' = v\sqrt{3}$.

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