Ball A is dropped from the top of a building 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 collsion occur?

Explanation

Given `V_(A)=2V_(B)`
Let `h_(1)"and"h_(2)` are the distances travelled by the two balls
`:. sqrt(2gh_(1))=2sqrt(u^(2)-2gh_(2))`
`2gh_(1)=4u^(2)-8gh_(32), 2gh_(1)+8gh_(2)=4u^(2), 2g[h_(1)+4h_(2)]=4u^(2)`
`h_(1)+4h_(2)=(2u^(2))/g` ...(1)
Again as `V_(A)=22V_(B),`from `v=u+at o+"gt" =2(u-"gt")`
`:."gt"=2u-2"gt" rArr 2u=3"gt"`
`:. t=2u//3g` ...(2)
`h_(1)+h_(2)=ut=u(2u/3g)=(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)`
`:. h_(1)/h_(2)=(2u^(2)//9g)/(4u^(2)//9g)=1/2`