question_answer

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)       \[2/5\]

Correct Answer: B

Solution :

[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) \[xh=ut-\frac{1}{2}g{{t}^{2}}\] or         \[xh=u\sqrt{\frac{2(1-x)h}{g}}-(1-x)h\] or         \[u=\sqrt{\frac{gh}{2(1-x)}}\]                 ??(3) Now  \[{{v}_{A}}=2{{v}_{B}}\] (at the time of collision) Or     \[v_{A}^{2}=4v_{B}^{2}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2g(1-x)h=4\left\{ {{u}^{2}}-2gxh \right\}\] or         \[2g(1-x)=4\left\{ \frac{gh}{2(1-x)}-2gxh \right\}\] or         \[(1-x)\,=\frac{1}{1-x}-4x\] or         \[1+{{x}^{2}}-2x=1-4x+4{{x}^{2}}\] or         \[x=\frac{2}{3}\]