A foot ball is kicked of with an initial speed of 19.6 m/sec at a projection angle `45^(@)`. A receiver on the goal line 67.4 m away in the direction of the kick starts running 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 2 theta)/(g) = ((19.6)^(2) xx sin 90)/(9.8)`
or `R = 39.2` metre.
Man must run 67.4m - 39.2 m = 28.2 m
in the time taken by the ball to come to ground.
Time taken by the ball.
`t = (2u sin theta)/(g) = (2 xx 19.6 xx sin 45^(@))/(9.8) = (4)/(sqrt(2))`
`t = 2sqrt(2) = 2 xx 1.41 = 2.82` sec.
Velocity of man `= (28.2m)/(2.82 sec) = 10m//sec`.