Two bodies are held separated by 9.8,m vertic | vedclass

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(C) Let the initial positions of the two bodies be $y_1 = 9.8\,m$ and $y_2 = 0\,m$.Both bodies are released simultaneously,so their initial velocity $

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$9.8$

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(C) Let the initial positions of the two bodies be $y_1 = 9.8\,m$ and $y_2 = 0\,m$.Both bodies are released simultaneously,so their initial velocity $u = 0\,m/s$ and acceleration $a = g$ (downwards).The position of the first body at time $t$ is given by $y_1(t) = 9.8 + 0(t) + \frac{1}{2}gt^2$.The position of the second body at time $t$ is given by $y_2(t) = 0 + 0(t) + \frac{1}{2}gt^2$.The relative distance between them is $|y_1(t) - y_2(t)| = |(9.8 + \frac{1}{2}gt^2) - (\frac{1}{2}gt^2)| = 9.8\,m$.Since both bodies experience the same acceleration due to gravity,their relative velocity and relative acceleration are both zero.Therefore,the distance between them remains constant at $9.8\,m$ regardless of the time elapsed.

Two bodies are held separated by $9.8\,m$ vertically one above the other. They are released simultaneously to fall freely under gravity. After $2\,s$,the relative distance between them is $............\,m$

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