Ball A is dropped from the top of a building and at the same instant that a ball B is thrown vertically upward 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)`
`impliesg t=2(u-g t)impliest=(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)`