A football is kicked at a speed of `20 m//s` a projection angle of `45^(@)`. A receiver on the goal line ` 25` meters away in the direction to the kick runs the same instant to meet the ball. Before it hits the ground?

Options

A. ` 5. 48 3 m//s`
B. `6. 28 3 m//s`
C. ` 5 . 112 m//s`
D. ` 6. 112 m//s`

Answer

A

Explanation

Here ` u = 20 m//s, theta = 54^@, d= 25 m` ltBrgt Horizontal range , ` R= (u^2 sin 2 theta) g` brgt ` R = (( 20)^2 sin 2 xx 45^@)/( 9.8) = (400)/(9.8) xx sin 90^@`
` = ( 400)/(9.8) = 40 . 82 m`
Time of flight, `
` T = (20 sin theta ) /g = ( 2 xx 20 sin 45)/( 9.8) = ( 2xx 20 xx 1)/( 9.8 xx sqrt 2)`
or ` T = 2 .886 s`.
Since the goal man is already ` 25 m` away in the direction of the ball So. to catch the ball, he is to cover a distacne ` = 40. 82 - 25 . 82 m`, in time ltBrgt ` 2. 886 sec`.
` :. the velocity of goal man to catch the ball is ` v= ( 15 . 82)/( 2.886) = 5 48 3 m//s`.