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 at time of collision velocity of `A=V_A`
`=2 times V_B(velocity of B)`
Let the body be dropped from a height `h`. Let the two stones collide at x from ground. For the body dropped,
`s=h-x=1/2g t ^2to (1)`
For the body thrown up,
`x=ut-1/2g t ^2to (2)`
For the body dropped,
`v=u+at impliesV_Ato(3)`
For the body thrown up,
`v=u-g t impliesV_B=u-g t to (4)`
Given `V_A=2V_B`
`impliesg t =2(u-g t )or u=(3 g t )/2to (5)`
Divide equation (1) with equation (2)
`(h-x)/x=(1/2g t ^2)/(ut-1/2g t ^2)impliesh/x-1=(1/2g t ^2)/(ut-1/2 g t^2)`
`impliesh/x(1/2g t^2+ut-1/2 g t ^2)/(ut-1/2g t ^2)`
`thereforeh/x=(ut)/(ut-1/2g t ^2)`
Put `u=(3g t )/2 ` then `h/x=(3g t^2//2)/(3 g t ^2//2-1/2g t ^2)`
`thereforeh/x=(3g t ^2)/(2g t ^2)=3/2thereforex=2/3h`
`therefore` Fraction of height of collision =`2/3`