Two bodies,one held 1 , m vertically above th | vedclass

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(A) Let the two bodies be $A$ and $B$. Body $A$ is $1 \, m$ above body $B$.Both bodies are released simultaneously and fall freely under gravity,meani

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(A) Let the two bodies be $A$ and $B$. Body $A$ is $1 \, m$ above body $B$.Both bodies are released simultaneously and fall freely under gravity,meaning their initial velocity $u = 0 \, m/s$ and acceleration $a = g = 9.8 \, m/s^2$ are the same for both.The position of body $A$ at time $t$ is given by $y_A = y_{A,0} + ut + \frac{1}{2}gt^2$.The position of body $B$ at time $t$ is given by $y_B = y_{B,0} + ut + \frac{1}{2}gt^2$.The relative separation between the bodies at any time $t$ is given by $|y_A - y_B| = |(y_{A,0} + ut + \frac{1}{2}gt^2) - (y_{B,0} + ut + \frac{1}{2}gt^2)|$.This simplifies to $|y_A - y_B| = |y_{A,0} - y_{B,0}|$.Since the initial vertical separation $|y_{A,0} - y_{B,0}|$ is $1 \, m$,the relative separation remains constant at $1 \, m$ regardless of the time elapsed,as both bodies experience the same acceleration due to gravity.

Two bodies,one held $1 \, m$ vertically above the other,are released simultaneously and fall freely under gravity. After $2 \, s$,the relative separation of the bodies will be: (in $, m$)

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