A football is kicked with speed ` 20 ms^(-1)` at a projection angle of ` 45^@` from the ground. A receiver on the goal line `20 m` away in the disrection of the kich runs the same instant to meet the ball. What must he his speed, of he is to catch the ball cefore it hits the ground ? Take ` g= 10 ms^(-2)`.

Explanation

Here, ` u= 20 ma^(-1) , theta= 45^@`,
Herizontal range,
` R = (u^2 sin 2 theta)/g = ( (20)^2 sin xx 45^@) /( 10) = 40 m`
Time of flight`, T = ( 2 u sin theta)/g = (2 xx 20 sin 45^@)/(10)`
` =4xx 1/ (sqrt 2) = 2 sqrt s`
distance required to be coverd by goal man to catch the ball ` = 40- 20 =20 m`.
Velocity of goal man to catch the ball is
` = (20) /(2 sqrt2)= 5 sqrt2 ms^(-1)`
` =5 xx 1. 414 = 7.070 ms^(-1)`.