Question

Ball A is dropped from the top of a building. At the same instant ball B is thrown vertically upwards from the
ground. When the balls collide, they are moving in opposite direction and the speed of A is twice the speed of
B. At what fraction of the height of the building did the collision occurs.

Answer 1

Solution

1/3

2/3

1/4

2/5

Answer :

B

Solution :

Let h be the total height and x the desired fraction. Initial velocity of ball B is u at time of collision it is vB. Then

(1−x)h=12gt2 …(1)

or t=2(1−x)hg−−−−−−−−−√ …(2)

Also, xh=ut−12gt2

or xh=u2(1−x)h−−−−−−−−√g−(1−x)h

or u=gh2(1−x)−−−−−−−−√

Now vA=2vB (at the time of collision)

or v2A=4v2B

:. 2g(1−x)h=4{u2−2gxh}

or 2g(1−x)h=4{gh2(2−x)−2gxh}

or (1−x)=11−x−4x

or 1+x2−2x=1−4x+4x2

or x=23.