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

$(1-x)h=\frac{1}{2}g{t}^2...(1)$

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

Also, $xh=ut-\frac{1}{2}g{t}^2$

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 $v_A^2=4v_B^2$

$\therefore 2g(1-x)h=4\{u^2-2gxh\}$

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 $x=\frac{2}{3}$