A ball A is dropped from the top of a building and at the same time an identical ball B is thrown vertically upward from the ground. When the balls collide the speed of A is twice that of B. At what fraction of the height of the building did the collison occur?

Explanation

Gicen `V_(A)=2V_(B)impliesgt=2(u-"gt")impliest=(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 height of the building where collision occur
is `(hB)/(H)=((4u^(2))/(9g))/((4u^(2))/(9g)+(2u^(2))/(9g))=(2)/(3)`