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 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?

The correct option is B $2/3$
Let $h$ be the total height and $x$ the desired fraction. Initial velocity of ball B is $u$ and at time of collision it is $v_B$. Then
Using second equation of motion
For ball A
$(1-x)h=\frac{1}{2}g{t}^2...\left(1\right)$
$\text{or}t=\sqrt{\frac{2(1-x)h}{g}}...\left(2\right)$
For ball B, we can write
$xh=u_{B}t-\frac{1}{2}g{t}^2$
By using equation (1) and (2) in above equation we get
$xh=u_B\sqrt{\frac{2(1-x)h}{g}}-(1-x)h$
$xh+h-xh=u_B\sqrt{\frac{2(1-x)h}{g}}$
$u_B=\sqrt{\frac{gh}{2(1-x)}}...\left(3\right)$
Now $v_A=2v_B$ (at the time of collision given)
or $v_A^2=4v_B^2$
By third equation of motion and for ball A
$u_A=0$ (Initial velocity )
$v_A^2=u_A^2+2gh$
Since $v_A^2=4v_B^2$
$\therefore 2g(1-x)h=4\{u_B^2-2gxh\}$
or $2g(1-x)h=4\{\frac{gh}{2(1-x)}-2gxh\}$
or $(1-x)=\frac{1}{(1-x)}-4x$
Multiply above equation by $(1-x)$
$(1-x{)}^2=1-4x(1-x)$
or $1+x^2-2x=1-4x+4x^2$
or $x=\frac{2}{3}$