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A train is moving along a straight line with a constant acceleration a. A body standing in the train throws a ball forward with a speed of `10ms^(-1)`, at an angle of `60^(@)` to the horizontal . The body has to move forward by 1.15 m inside the train to cathc the ball back to the initial height. the acceleration of the train. in `ms^(-2)` , is: |
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Answer» Correct Answer - 5 `T=(2u sin theta)/(g)=(2xx10xxsqrt(3))/(10xx2)=sqrt(3)s` `R=u cos thetaT-(1)/(2)aT^(2)` `1.15=10x(1)/(2)sqrt(3)-(1)/(3)a(sqrt(3))^(2)` `(3)/(2)a=5sqrt(3)-1.15=8.65-1.15=7.5` `a=7.5xx(2)/(3)=6ms^(-2)` |
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