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(D) Let the first body be released at $t = 0$. Its position after $t$ seconds is given by $y_1 = \frac{1}{2}gt^2$.The second body is released at $t =

Vedclass · Accepted Answer

$24.5$

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(D) Let the first body be released at $t = 0$. Its position after $t$ seconds is given by $y_1 = \frac{1}{2}gt^2$.The second body is released at $t = 1 \; s$. Its position after $t$ seconds (where $t > 1$) is given by $y_2 = \frac{1}{2}g(t - 1)^2$.We need to find the separation $2 \; s$ after the release of the second body. This corresponds to a total time of $t = 1 + 2 = 3 \; s$ from the release of the first body.Position of the first body at $t = 3 \; s$: $y_1 = \frac{1}{2} 	imes 9.8 	imes (3)^2 = 4.9 	imes 9 = 44.1 \; m$.Position of the second body at $t = 3 \; s$ (which is $2 \; s$ after its release): $y_2 = \frac{1}{2} 	imes 9.8 	imes (2)^2 = 4.9 	imes 4 = 19.6 \; m$.Separation between the two bodies: $\Delta y = y_1 - y_2 = 44.1 - 19.6 = 24.5 \; m$.

$A$ body is released from a great height and falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies,two seconds after the release of the second body is.......$m$

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