Question

A body $A$ is thrown vertically upwards with such a velocity that it reaches a maximum height of $h$. Simultaneously, another body $B$ is dropped from height $h$. It strikes the ground and does not rebound. The relative velocity of $A$ wrt $B$ to time graph is best represented by : (upward direction is positive)

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

As per the question,
the initial velocity of the body $A$ is given by $u_A=\sqrt{2gh}$
For the first half when the ball $A$ goes up
$V_A=u_A-gt$, where $u_A=\sqrt{2gh}$
and,
$V_B=-gt$ (As it is dropped so, $u_B=0$)
$\therefore$ $V_{AB}=V_A-V_B=u_A=\sqrt{2gh}$.....(i)
So, $V_{AB}$ is constant

Now, for the second half, when ball comes down
$V_A=-gt$
and,
$V_B=0$ (as it did't rebound)
So, $V_{AB}=V_A-V_B=-gt$....(ii)

Hence, from equation $\left(i\right)$ we observe that relative velocity is independent of $t$, which will be valid till ball $A$ reaches maximum height. Hence for the time $t$, $V_{AB}$ will be constant. Whereas after time $t$, relative velocity changes its direction and increases afterwards.