A player kicks a football at an angle of 30^{ | vedclass

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Question

(C) The total range $R$ of the projectile is given by:$R = \frac{u^2 \sin 2	heta}{g} = \frac{30^2 \sin(2 	imes 30^{\circ})}{10} = \frac{900 	imes \

Vedclass · Accepted Answer

$8 \sqrt{3} \,ms^{-1}$

Vedclass · Answer

(C) The total range $R$ of the projectile is given by:$R = \frac{u^2 \sin 2	heta}{g} = \frac{30^2 \sin(2 	imes 30^{\circ})}{10} = \frac{900 	imes \sin 60^{\circ}}{10} = 90 	imes \frac{\sqrt{3}}{2} = 45 \sqrt{3} \,m$.The time of flight $T$ is given by:$T = \frac{2u \sin 	heta}{g} = \frac{2 	imes 30 	imes \sin 30^{\circ}}{10} = \frac{60 	imes 0.5}{10} = 3 \,s$.The second player is standing at a distance $d = 21 \sqrt{3} \,m$ from the first player. To catch the ball before it hits the ground, the player must reach the point where the ball lands (or any point along its trajectory). The minimum speed is required to reach the landing point $R$ within the time of flight $T$.The distance the second player needs to cover is $S = R - d = 45 \sqrt{3} - 21 \sqrt{3} = 24 \sqrt{3} \,m$.Since the player starts at the same instant as the kick, the time available to cover this distance is the time of flight $T = 3 \,s$.Therefore, the minimum speed $v$ is:$v = \frac{S}{T} = \frac{24 \sqrt{3}}{3} = 8 \sqrt{3} \,ms^{-1}$.Thus, the correct option is $C$.

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