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 just before it reaches the ground, he should run with a velocity equal to (Take g= 10 $ms-2$)

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

Answer 1

The correct option is B 4$\sqrt{2}$ $m{s}^{-1}$

Horizontal range is maximum When angle of projection $\theta$ is 45${}^{\circ }$

$R_{max}$= $\frac{u^2}{g}$= $\frac{{20}^2}{10}$= 40m
Time of flight, $T=\frac{2usin{45}^{\circ }}{g}=\frac{2\times 20\times \frac{1}{\sqrt{2}}}{10}=2\sqrt{2}$s

The player can catch the ball before reaching the ground if he covers a distance $=40-24=16min2\sqrt{2}$s

The velocity of that player must be $=\frac{16}{2\sqrt{2}}=4\sqrt{2}m{s}^{-1}$