In the figure the top view of a compartment of a train is shown. A man is sitting at a corner `'B'` of the compartment . The man throws a ball `(` with respect to himself `)` along the surface of the floor towards the corner `'D'` of the compartment of the train. The ball hits the corner `'A'` of the compartment, then find the time at which it hits `A` after the ball is thrown . Assume no other collision during motion and floor is smooth. The length of the compartment is given `'l'` and the train is moving with constant acceleration `'a'` in the direction shown in the figure.

Answer

`t=sqrt((2l)/(a))`

Explanation

Solving the problem in the frame of train. Taking origin as corner `'B'`.
Along `x` axis `x-`
`x=ucos theta t` ………`(1)`
Along `y` axis `y-`
`y=u_(y)+(1)/(2)a_(y)t^(2)`

`0=u sintheta t-(1)/(2)a t^(2)`.......`(2)`
As the ball is thrown towards `'D'`
`tan theta=(l)/(x)` .......`(3)`
From equation `(1),(2)& (3) ` we get
`t=sqrt((2l)/(a))` required time after which ball hit the corner.