A player kicks a ball at a speed of 20 ms^{- | vedclass

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(B) For maximum horizontal range,the angle of projection is $	heta = 45^{\circ}$.The maximum horizontal range is given by $R_{\max} = \frac{u^2}{g} =

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$4 \sqrt{2} \ ms^{-1}$

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(B) For maximum horizontal range,the angle of projection is $	heta = 45^{\circ}$.The maximum horizontal range is given by $R_{\max} = \frac{u^2}{g} = \frac{(20)^2}{10} = 40 \ m$.The time of flight $T$ is given by $T = \frac{2u \sin 	heta}{g} = \frac{2 	imes 20 	imes \sin 45^{\circ}}{10} = 4 	imes \frac{1}{\sqrt{2}} = 2 \sqrt{2} \ s$.The second player is $24 \ m$ away from the starting point. The ball lands at a distance of $40 \ m$ from the starting point.Therefore,the distance the second player needs to cover is $x_2 = R_{\max} - 24 = 40 - 24 = 16 \ m$.The velocity $V$ of the second player required to reach this point in time $T$ is $V = \frac{x_2}{T} = \frac{16}{2 \sqrt{2}} = \frac{8}{\sqrt{2}} = 4 \sqrt{2} \ ms^{-1}$.

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