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A train is moving along a straight line ...

A train is moving along a straight line with a constant acceleration 'a' . A boy standing in the train throws a ball forward with a speed of `10 m//s `, at an angle of `60(@)` to the horizontal. The boy has to move forward by `1.15 m ` inside the train to catch the ball back at the initial height . the acceleration of the train , in `m//s^(2)` , is

Text Solution

Verified by Experts

The correct Answer is:
5

` T = (2usintheta)/g = (2xx10xxsqrt(3))/g = (2xx10xxsqrt(3))/(10xx2) = sqrt(3)s`.
`R = u cos theta T - 1/2 aT^2`
`1.15 = 10 xx 1/2 sqrt(3) - 1/2 a (sqrt3)^2`
`3/2 a = 5 sqrt(3) - 1.15 = 8.65 - 1.15 = 7.5` ,brgt ` a = 7.5 xx 2/3 = 5ms^(-2)`.
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