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(C) Let the first stone fall for time $t_0$. Its distance covered is $S_1 = \frac{1}{2} g t_0^2$. The second stone starts $\Delta t$ seconds later,so

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$\frac{H}{g \Delta t} + \frac{\Delta t}{2}$

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(C) Let the first stone fall for time $t_0$. Its distance covered is $S_1 = \frac{1}{2} g t_0^2$. The second stone starts $\Delta t$ seconds later,so it falls for time $(t_0 - \Delta t)$. Its distance covered is $S_2 = \frac{1}{2} g (t_0 - \Delta t)^2$. The separation distance $H$ is given by $H = S_1 - S_2$. Substituting the expressions: $H = \frac{1}{2} g t_0^2 - \frac{1}{2} g (t_0 - \Delta t)^2$. Expanding the term: $H = \frac{1}{2} g [t_0^2 - (t_0^2 - 2 t_0 \Delta t + \Delta t^2)]$. $H = \frac{1}{2} g [2 t_0 \Delta t - \Delta t^2]$. $H = g t_0 \Delta t - \frac{1}{2} g \Delta t^2$. Rearranging for $t_0$: $g t_0 \Delta t = H + \frac{1}{2} g \Delta t^2$. Dividing by $g \Delta t$: $t_0 = \frac{H}{g \Delta t} + \frac{\Delta t}{2}$.

Two stones begin to fall from rest from the same height,with the second stone starting to fall $\Delta t$ seconds after the first. The distance of separation between the two stones becomes $H$,$t_0$ seconds after the first stone starts its motion. Then $t_0$ is equal to:

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