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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 minside the train to catch the ball back at the initial height . the acceleration of the train , in m//s^(2) , is

Answer» <html><body><p>3 m `s^(-2)` <br/>5 m `s^(-2)` <br/><a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a> m `s^(-2)` <br/><a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a> m `s^(-2)` </p>Solution :Here, `u=10ms^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>),theta=60^(@)` <br/> `:.u_(x)=ucostheta=10xxcos60^(@)=<a href="https://interviewquestions.tuteehub.com/tag/5ms-1900301" style="font-weight:bold;" target="_blank" title="Click to know more about 5MS">5MS</a>^(-1)` and `u_(y)=usintheta=10xxsin60^(@)=5sqrt3ms^(-1)` <br/> `:.t=(2u_(y))/(<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>)=(2xx5sqrt3)/(10)=sqrt(3)s` <br/> and `1.15=u_(x)t-(1)/(2)at^(2),` where a is the acceleration of train <br/> `1.15=5xxsqrt(3)-(1)/(2)xxaxx(sqrt(3))^(2)` <br/> `(3a)/(2)=5sqrt(3)-1.15=8.65-1.15-7.5` <br/> `a=7.5xx(2)/(3)=5ms^(-2)`</body></html>


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