Ball A is dropped from the top ofa 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?

Ball A is dropped from the top ofa building at the same instant that a

1.	Ball A is dropped from the top ofa 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?
Answer» Solution :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`