A cricket ball is thrown at a speed of ` 28 ms^(-1)` in a direction `30^(@)` above the horizontal . Calculate (a) the maximum height , (b) the time taken by the ball to return to the same thrower to the point where the ball returns to the same level.

Explanation

The maximum height is given by
`h_(m)=((v_(0)sintheta_(@))^(2))/(2g)=((28sin30^(@))^(2))/(2(9.8))m`
`=(14xx14)/(2xx9.8)=10.0m`
(b) The time taken to return to the same level is `T_(f)=(2v_(@)sintheta_(@))//g=(2xx28xxsin30^(@))//9.8=28//9.8s=2.9s`
(c ) The distance from the thrower to the point where the ball returns to the same level is
`R=((v_(@)^(2)sin2theta_(@)))/(g)=(28xx28xxsin60^(@))/(9.8)=69m`