In a cricket match, a batsman hits the ball in air. A fielder, originally standing at a distance of 12 m due east of the batsman, starts running 0.6 s after the ball is hits. He runs towards north at a constant speed of `5 m//s` and just manages to catch the ball 2.4 s after he starts running. Assume that the ball was hit and caught at the same height and take `g = 10 m//s^(2) g = 10 m//s^(2)` Find the speed at which the ball left the bat and the angle that its velocity made with the vertical.

Answer

`u = 16 m//s; theta = tan^(-1) (4sqrt(2))/(15)`