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 ball to return ti the same level, and (c )the distance from the thrower to the point wher the ball restance from the throwrt to the point where the ball returns to the same level.

Explanation

Initial velocity , `u=28m//s`
Angle of projection , `theta=30^(@)`
(a) maximum height ,
`H=(u^(2) sin ^(2)theta)/(2g)=(28^(2)(sin^(2)30^(@))/(2xx9.8))=10 m`
(b) Time taken by the ball to return to the same level
`T=(2u sin theta)/(g)=(2xx28xxsin 30^(@))/(9.8)=2.9s`
(c) The diatance from the thrower to the point where the ball return to the same level=horizontal range covered by the ball .
`R=(u^(2) sin 2 theta)/(g)=(28^(2) sin 60^(@))/(9.8)=69.3m`