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

A train is moving along a straight line with a constant acceleration '

1.	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» 3 m `s^(-2)` 5 m `s^(-2)` 8 m `s^(-2)` 6 m `s^(-2)` Solution :Here, `u=10ms^(-1),theta=60^(@)` `:.u_(x)=ucostheta=10xxcos60^(@)=5MS^(-1)` and `u_(y)=usintheta=10xxsin60^(@)=5sqrt3ms^(-1)` `:.t=(2u_(y))/(G)=(2xx5sqrt3)/(10)=sqrt(3)s` and `1.15=u_(x)t-(1)/(2)at^(2),` where a is the acceleration of train `1.15=5xxsqrt(3)-(1)/(2)xxaxx(sqrt(3))^(2)` `(3a)/(2)=5sqrt(3)-1.15=8.65-1.15-7.5` `a=7.5xx(2)/(3)=5ms^(-2)`