Relative Motion in One Dimension PYQ | JEE Main Physics 2025

 

❓ Question

A particle A is projected vertically upward from the ground with speed $u$. At the same instant, another particle B is released from rest from the top of a tower of height $h$.

Find the distance $d$ between the two particles after time $t$.

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✍️ Solution

Initially, the separation between the particles is

$$h.$$

After time $t$,

$$h=h_1+h_2+d,$$

where

• $h_1$ = upward displacement of particle $A$,
• $h_2$ = downward displacement of particle $B$,
• $d$ = distance between the particles.

Hence,

$$d=h-(h_1+h_2).$$

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Motion of Particle $A$

Taking upward as positive,

$$u=u,\qquad a=-g.$$

Using

$$s=ut+\frac12at^2,$$

we get

$$h_1=ut-\frac12gt^2.$$

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Motion of Particle $B$

Since particle $B$ is released from rest,

$$u=0,\qquad a=+g.$$

Therefore,

$$h_2=\frac12gt^2.$$

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Distance Between the Particles

Substitute $h_1$ and $h_2$:

$$\begin{aligned} d &=h-\left(h_1+h_2\right)\\ &=h-\left(ut-\frac12gt^2+\frac12gt^2\right)\\ &=h-ut. \end{aligned}$$

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✅ Final Answer

$$\boxed{d=h-ut}$$