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:

Explanation

From the perspective of observer `A` , considering vertical motion of the ball from the point of throw till it reaches back at the initial height .
` U_(y) = + 5 sqrt(3) m//s , S_(y) = 0 , a_(y) = -10 m//s^(2) , t = ? `
Applying ` S = ut+ 1/2 at^(2) rArr 0 = 5 sqrt((3)t - 5 t^(2)`
`:. t = sqrt(3) sec`
Considering horizontal motion from the persective of observer `B` . let `u` be the speed of train at the time of throw .
The horizontal distance travelled by the ball `= ( u+5 )sqrt(3)`.
The horizontal distance travelled by the boy
`= [ u sqrt(3) + 1/2 a(sqrt(3))^(2)] + 1.15`
As the boy catches the ball therefore
`:. (u+5)sqrt(3) = u sqrt(3) + 3/2 a + 1.15`
`:. 5 sqrt(3) = 1.5 a + 1.15 ` `:. 7.51 = 1.5 a `
`:. a ~~ 5 m//s^(2)`