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

Let height of the building = H
Let two balls collide at a height = h
For ball A, u = 0, V = VA s = h, t = t, a = g
Substituting these values in `s=ut+1/2at^(2)`
`H-h=0+1/2"gt"^(2)`
`H-h=1/28t^(2)" ...(1)"`
and `V_(A)="gt ...(2)"`
For ball B, u = u, v = `v_(B)`, s = h, a = -g
Substituting these values in `s=ut+1/2at^(2)`
`rArr h=ut-1/2"gt"^(2)" ...(3)"`
and `V_(B)=u-"gt .... (4)"`
given `V_(A)=2V_(B)`
`"gt"=2(u-"gt")`
`u=3/2"gt .....(5)"`
`((1))/((2)) rArr (H-h)/h=(1/2"gt"^(2))/(ut-1/2"gt"^(2))`
`=(1/2"gt"^(2))/(3/2"gt"^(2)-1/2"gt"^(2))` [`because` from (5)]
`(H-h)/h=1/2 rArr H/h-1=1/2 rArr H/h=1+1/2=3/2`
`h/H=2/3`