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 gt =2(u-gt) rArr t=(2u)/(3g)`
Displacement of A is `S_(A)=1/2 gt^(2) =(2u^(2))/(9g)`
Displacement of B is `s_B=ut-1/2 gt^(2)=(4u^(2))/(9g)`
Now fraction of hightt of the building where condition occur is `h_B/H=((4u^(2))/(9g))/((4u^(2))/(9g)+(2u^(2))/(9g))=2/3`