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 level, and (c )the distance from the 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`