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A cricket ball is thrown at a speed ofms^(-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. |
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Answer» Solution :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` |
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