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 ?

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 collision occur ?
Answer» Solution :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`