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 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?

• A. 1/3
• B. 2/3
• C. 1/4
• D. 2/5

Answer 1

Answer: (B)

Let $h$ be the total height and $x$ the desired fraction. Initial velocity of ball $B$ is $u$ at time of collision it is $v_B$. Then
$(1 - x) h = \frac{1}{2} gt^2 \,...(1)$
or $t = \sqrt{\frac{2(1 - x)h}{g}} \,...(2)$
Also, $xh = ut - \frac{1}{2} gt^2$
or $ xh = u\sqrt{\frac{2(1 - x)h}{g}} - ( 1 - x)h$
or $u = \sqrt{\frac{gh}{2(1 - x)}} \,...(3)$
Now $v_A = 2v_B$ (at the time of collision)
or $v_A^2 = 4 v_B^2$
$\therefore 2g ( 1 - x) h = 4 \{u^2 - 2gxh\}$
or $2g( 1 - x)h = 4\{ \frac{gh}{2(1 - x)} - 2gxh \}$
or $( 1 - x) = \frac{1}{1 - x} - 4x$
or $ 1 + x^2 - 2x = 1 - 4x + 4x^2$
or $ x = \frac{2}{3}$