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

Given `V_(A)=2V_(B)`
Let `h_(1)` and `h_(2)` are the distances travelled by the two balls
`therefore sqrt(2gh_(1))=2sqrt(u^(2)-2gh_(2))`
`2gh_(1)=4u^(2)-8gh_(2),2gh_(1)+8gh_(2)=4u^(2),2g[h_(1)+4h_(2)]=4u^(1)`
`h_(1)+4h_(2)=(2u^(2))/(g)`.........(1)
Again as `V_(A)=2V_(B)`,from v=u+at o +gt=2 (u-gt)
`therefore` gt=2u-2gt`implies`2u=3gt
`therefore t=(2u^(2))/(3g)` ..........(3)
`h_(1)+4h_(2)=(2u^(2))/(g)`.........(4)
Solving (3) & (4) we get `h_(1)(2u^(2))/(9g)" ",h_(2)=(4u^(2))/(9g)`
`therefore (h_(1))/(h_(2))=(2u^(2)//9g)/(4u^(2)//9g)=(1)/(2)`