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A body released from a great height falls freely towards the earth. Another body is released from the same height exactly a second later. Then the separation between two bodies, $2 \mathrm{~s}$ after the release of the second body is nearly
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$25 \mathrm{~m}$
Let the first body fall through a height $h_1$ during $3 \mathrm{~s}$
$\therefore \quad h_1=\frac{1}{2} g(3)^2=\frac{9}{2} g$ and the second body falls through a height $h_2$ during $2 \mathrm{~s}$ $\begin{aligned} & h_2=\frac{1}{2} g(2)^2=\frac{4}{2} g \quad \therefore h_1-h_2 \\ & =\frac{9}{2} g-\frac{4}{2} g=\frac{5}{2} g=25 \mathrm{~m}\end{aligned}$
$\therefore \quad h_1=\frac{1}{2} g(3)^2=\frac{9}{2} g$ and the second body falls through a height $h_2$ during $2 \mathrm{~s}$ $\begin{aligned} & h_2=\frac{1}{2} g(2)^2=\frac{4}{2} g \quad \therefore h_1-h_2 \\ & =\frac{9}{2} g-\frac{4}{2} g=\frac{5}{2} g=25 \mathrm{~m}\end{aligned}$
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