Ball `A` dropped from the top of a building. A the same instant ball `B` is thrown vertically upwards 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

`(u^(2)h)/(u^(2)+2gh)`

Explanation

We are given the ratio of velocity of `A` to the velocity of `B` at the point of collision is same as the ration of height of this point form top of the building to the heitht height from the groun, let this ratio be `k`.
form figure `(v_(A))/(v_(B))=((1-x)h)/(xh)/(x) =k`
Hence, `v_(A)=k v_(B)`
`rArr sqrt (2g(1-x)h)=k sqrt((u^(2)-2gxh)`
Hence, `2g(1-x)h=k^(2)(u^(2)-2gxh)`
`2g ((1+x))/(x)g =k^(2) ((u^(2))/(x)-2 gh)`
`2gkh =k^(2)(u^(2)/(x-2gh)`
`k=(2ghx)/((u^(2)-2hgx))`
`((1-x))/(x) =(A)/((u^(2)-A)) (let `2ghx=A) rArr z=(u^(2)-2A)/(u^(2))`.
`rArr x (1+(2gh)/(u^(2)))=1` or `x=u^(2)/((u^(2)+2gh)`
Hence , `xh =(u^(2)h)/(u^(2)+2gh)`
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