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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 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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The correct Answer is:
30
`u^(2) sin^(2) (45^(@))/(2(10))=120 rArr u^(2)=4800`
After collision, `(v_(y))/(v_(x))=tan 30^(@) rArr v_(x)=sqrt3v_y`
Also, `1/2mv^(2)=1/2(1/2mu^(2)) rArr v^(2)=(4800)/(2)=2400`
`v_(x)^(2)+v_(y)^(2)=2400 rArr 3v_(y)^(2)+v_(y)^(2)=2400 rArr v_(y)^(2)=600, H_(2)=(v_(y)^(2))/(2g)=(600)/(20)=30m`
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