Ball `A` dropped from the top of a building. A the same instant ball `B` is thrown vertically upwards from the ground. When the balls collide, they are moving in opposite directions and the speed of `A` is twice the speed of `B`. At what fraction of the height of the building did the collision occur?

Explanation

Given `V_A=2V_B`
`rArr g t=2(u- g t) rArr t=(2u)/(3g)`
Displacement of A is `S_A=1/2 g t^2 =(2u^2)/(9g)`
Displacement of B is `S_B=ut -1/2 g t^2 =(4u^2)/(9g)`
Now fraction of hight of the building where collision occur is
`h_B/H=((4u^2)/(9g))/((4u^2)/(9g)+(2u^2)/(9g))=2/3`