At the same instant, ball `A` is ball `B` is projected vertically building of height `h` and ball `B` is projected vertically upward from the ground with velocity `u`. The ration of velocty of `A` to the velocity of `B` at the point of cllistion is same asthe ratio of height of this point from top of the building to the height from the ground, find the height of the point of collision above theground.

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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