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

Step-by-step solution

aligned t & =T= 2 u g & = 2 10 60^ 10 = 3 ~s aligned Displacement of train in time t= 1 2 a t^2 Displacement of boy with respect to train =1.15 ~m Displacement of boy with respect to ground = (1.15+ 1 2 a t^2 ) Displacement of ball with respect to ground = (u 60^ ) t To catch the ball back at initial height, aligned 1.15+ 1 2 a t^2 & = (u 60^ ) t 1.15+ 1 2 a( 3 )^2 & =10 1 2 3 aligned Solving this equation, we get a=5 ~ms ⁻² Answer is 5 . Analysis of Question (i) Question is moderately tough. (ii) Velocity of ball