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(C) Let the first body fall for time $n$. Its displacement is $y_1 = \frac{1}{2} g n^2$.The second body starts after a time interval $N$,so it falls f

Vedclass · Accepted Answer

$\frac{1}{gN} + \frac{N}{2}$

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(C) Let the first body fall for time $n$. Its displacement is $y_1 = \frac{1}{2} g n^2$.The second body starts after a time interval $N$,so it falls for time $(n - N)$. Its displacement is $y_2 = \frac{1}{2} g (n - N)^2$.The vertical separation between the two bodies is given as $1 \, m$,so $y_1 - y_2 = 1$.Substituting the expressions: $\frac{1}{2} g n^2 - \frac{1}{2} g (n - N)^2 = 1$.Expanding the term $(n - N)^2 = n^2 - 2nN + N^2$:$\frac{g}{2} [n^2 - (n^2 - 2nN + N^2)] = 1$.$\frac{g}{2} [2nN - N^2] = 1$.$gNn - \frac{gN^2}{2} = 1$.$gNn = 1 + \frac{gN^2}{2}$.Dividing by $gN$: $n = \frac{1}{gN} + \frac{N}{2}$.

Two bodies begin a free fall from the same height at a time interval of $N \, s$. If the vertical separation between the two bodies is $1 \, m$ after $n \, s$ from the start of the first body,then $n$ is equal to:

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