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 collisionoccur?

Ball A is dropped from the top of a building and at the same instant t

1.	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 collisionoccur?
Answer» Solution :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)`