A player kicks a foot ball at a speed of $20 m s^{- 1}$ so that its horizontal range is maximum. Another player $24$ m away in the direction of kick starts running in the same direction at the same instant of hit, if he has to catch the ball just before it reaches to ground, he should run with a velocity equal to ( $g = 10 m s^{- 2}$ )

4 m s^{- 1}

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5.65 m s^{- 1}

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10 m s^{- 1}

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12 m s^{- 1}

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Solution

The correct option is B $5.65 m s^{- 1}$
Maximum range of ball $\Rightarrow R = \frac{u^{2} sin 2 θ}{g}$
Now, $R = \frac{20 \times 20 \times 1}{10} = 40 m$
Distance travelled by second man= $40 - 24 = 16 m$
Time= $t = \frac{40}{u cos 45^{o}} = 2 \sqrt{2} S$
Speed $v = \frac{d i s t a n c e}{t i m e} = \frac{16}{2 \sqrt{2}} = 5.65 m / s$

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