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 occurs?

Let h be the total height and x the desired fraction. Initial velocity of ball B is u at time of collision it is ${\mathrm{v}}_{\mathrm{B}}. \mathrm{Then}$

$(1-\mathrm{x})\mathrm{h} = \frac{1}{2}{\mathrm{gt}}^2------\left(1\right)$

or $\mathrm{t} = \sqrt{\frac{2(1-\mathrm{x})\mathrm{h}}{\mathrm{g}}} ------\left(2\right)$

Also, xh = ut-$\frac{1}{2}{\mathrm{gt}}^2$

or xh=u$\sqrt{\frac{2(1-\mathrm{x})\mathrm{h}}{\mathrm{g}}}-(1-\mathrm{x})\mathrm{h}$

or u = $\sqrt{\frac{\mathrm{gh}}{2(1-\mathrm{x})}}$

Now ${\mathrm{v}}_{\mathrm{A}} = 2{\mathrm{v}}_{\mathrm{B}} (\mathrm{at} \mathrm{the} \mathrm{time} \mathrm{of} \mathrm{collision})$

or ${{\mathrm{v}}_{\mathrm{A}}}^2 = 4{{\mathrm{v}}_{\mathrm{B}}}^2$

$\therefore 2\mathrm{g}(1-\mathrm{x})\mathrm{h} = 4\{\frac{\mathrm{gh}}{2(1-\mathrm{x})}-2\mathrm{gxh}\}$

or (1-x) = $\frac{1}{1-\mathrm{x}}-4\mathrm{x}$

or $1+{\mathrm{x}}^2-2\mathrm{x} = 1-4\mathrm{x}+4{\mathrm{x}}^2$

or x = $\frac{2}{3}$