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A player kicks a foot ball obliquely at a speed of $20 {ms}^{- 1}$ so that its range is maximum. Another player at a distance of 24m away in the direction of kick starts running at that instant to catch the ball. Before the ball hits the ground to catch it, the speed with which the second player has to run is $(g = 10 {ms}^{- 2})$

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4 {ms}^{- 1}

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

8 \sqrt{2} {ms}^{- 1}

8 {ms}^{- 1}

answer is B.

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Detailed Solution

$u = 20 m/s$

$R_{\max} = \frac{u^{2}}{g}$

$= \frac{20 \times 20}{10} = 40 m$

$T = \frac{2 u \sin θ}{g}$

$= \frac{2 \times 20 \times \frac{1}{\sqrt{2}}}{10} = \frac{4}{\sqrt{2}}$

$v = \frac{S}{T} = \frac{16}{\frac{4}{\sqrt{2}}} = 4 \sqrt{2} m/s$

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