A player kicks a ball at a speed of `20ms^(-1)` so that its horizontal range is maximum. Another players 24m 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 equl to `(take g=10ms^(-2))`

Options

A. `2sqrt(2) ms^(-1)`
B. `4sqrt(2) ms^(-1)`
C. `6sqrt(2) ms^(-1)`
D. `10sqrt(2)ms^(-1)`

Answer

B

Explanation

Horizontal range, `R=(u^(2)sin2theta)/(g)`
Horizontal range is maximum when angle of projection `theta" is "45^(@)`
`R_(max)=(u^(2))/(g)=(20^(2))/(10)=40m`
Time of flight, `T=(2usin45^(@))/(g)=(2xx20xx(1)/(sqrt(2)))/(10)=2sqrt2s`
The player can catch the ball before reaching the ground if he covers a distance `=40-24=16m" in "2sqrt(2)s`. The velocity of that plyare must be `=(16)/(2sqrt(2))=4sqrt(2)ms^(-1)`