A football is kicked with 20m/s at a projection angle of `45^(@)`. A reciever on the goal line 45 metres away in the direction o thekick runs the same instant to meet the ball. What must be his speed, if he is to catch the ball before it hits ground?

Explanation

Here `u=20m//s,theta=45^(@),d=25m`
Horizontal range `R=(u^(2))/g sin 2 theta=(20^(2))/9.8 sin 2 xx45^(@)=40.82m`
Time of flight `T=(2usin theta)/g=(2xx20)/9.8xxsin 45^(@)=2.886s`
Since the goal man is already 25 m away in the direction of the ball so to catc the ball, he is to cover a distance =40.82-25=15.82m, in time 2.886s. Therefore, the velocity of goal man to catch the ball is `v=15.82/2.886=5.486ms^(-1)`