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

$(a)\;\frac{1}{3} \quad (b)\;\frac{2}{3} \quad (c)\;\frac{1}{4}\quad (d)\;\frac{2}{5}$

Let h be total height let x be desired fraction.
For Ball A
$(1-x)h=\large\frac{1}{2} $$gt^2 =>t=\sqrt {\large\frac{2(1-x)h}{g}} For Ball B xh=ut-\large\frac{1}{2}$$ gt^2$
$=>xh=u\bigg(\sqrt {\large\frac{2(1-x)h}{g}}\bigg)$$-(1-x)h Solving u=\sqrt {\large\frac{gh}{2(1-x)}} v_A=2v_B at the time of collision {v_A}^2=4 {v_B}^2 2g(1-x)h=4 \{u^2-2gxh\} (1-x)=\large\frac{1}{1-x}$$-4x$
$x=\large\frac{2}{3}$
Hence b is the correct answer.
edited May 24, 2014
Nice and easy solution.