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Ball A dropped from the top of a buildin...

Ball `A` dropped from the top of a building. A the same instant ball `B` is thrown vertically upwards 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 collision occur?

Text Solution

Verified by Experts

Gicen `V_(A)=2V_(B)impliesgt=2(u-"gt")impliest=(2u)/(3g)`
Displacement of A is `S_(A)=(1)/(2)"gt"^(2)=(2u^(2))/(9g)`
Displacement of B is `S_(B)=ut-(1)/(2)"gt"^(2)=(4u^(2))?(9g)`
Now fraction of height of the building where collision occur
is `(hB)/(H)=((4u^(2))/(9g))/((4u^(2))/(9g)+(2u^(2))/(9g))=(2)/(3)`
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