A foot ball is kicked off with an initial speed of `19.6m//s` to have maximum range. Goal keeper standing on the goal line `67.4 m` away in the direction of the kick starts running opposite to the direction of kick to meet the ball at that instant. What must his speed be if he is to catch the ball before it hits the ground?

Explanation

`R=(u^(2)sin 2theta)/g=((19.6)^(2)sin90)/9.8`
or `R=39.2` metre.
Man must run `67.4 m-39.2m=28.2m` in the time taken by the ball to come to ground Time taken by the ball.
`t=(2u sin theta)/g=(2xx19.6xxsin45^(@))/9.8=4/sqrt2`
`t=2sqrt2=2xx1.41=2.82sec`.
Velocity of man`=(28.2m)/(2.82sec)=10m//sec`.