Question

A player kicks a 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 jusy before it reaches the ground, he should run with a velocity equal to

(Take g = 10 m s^-1)

• A. $2\sqrt{2}ms^{- 1}$
• B. $4\sqrt{2}ms^{- 1}$
• C. $6\sqrt{2}ms^{- 1}$
• D. $10\sqrt{2}ms^{- 1}$

Answer 1

Answer: (B)

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

Answer 2

Horizontal range, $R = \frac{u^{2}\sin 2\theta}{g}$

Horizontal range is maximum when angle of projections $\theta$is $45{^\circ}$,

${R\frac{u^{2}}{g}\frac{20^{2}}{10}}_{\max}$

Time of flight,$T = \frac{2u\sin 45{^\circ}}{g} = \frac{2 \times 20 \times \frac{1}{\sqrt{2}}}{10} = 2\sqrt{2}s$

The player can catch the ball before reaching the ground if he covers a distance $= 40 - 24 = 16m$in $2\sqrt{2}s.$ The velocity of that player must $= \frac{16}{2\sqrt{2}} = 4\sqrt{2}ms^{- 1}$