To solve this problem, we need to determine the time, denoted as $ \mathrm{t}_0 $, at which the separation between two falling stones becomes $ \mathrm{H} $. The second stone starts falling $ \Delta \mathrm{t} $ seconds after the first stone.

Distance Fallen by the First Stone ($ \mathrm{S}_1 $):

The first stone falls for $ \mathrm{t}_0 $ seconds, so the distance it covers is given by:

$ \mathrm{S}_1 = \frac{1}{2} \mathrm{~g} \mathrm{t}_0^2 $

Distance Fallen by the Second Stone ($ \mathrm{S}_2 $):

The second stone starts falling $ \Delta \mathrm{t} $ seconds after the first, so it falls for $ (\mathrm{t}_0 - \Delta \mathrm{t}) $ seconds. Thus, the distance it covers is:

$ \mathrm{~S}_2 = \frac{1}{2} \mathrm{~g} (\mathrm{t}_0 - \Delta \mathrm{t})^2 $

Distance of Separation ($ \mathrm{H} $):

The separation distance between the two stones is:

$ \mathrm{H} = \mathrm{S}_1 - \mathrm{S}_2 = \frac{1}{2} \mathrm{~g} \mathrm{t}_0^2 - \frac{1}{2} \mathrm{~g} (\mathrm{t}_0 - \Delta \mathrm{t})^2 $

Simplifying the equation:

$ \begin{aligned} \mathrm{H} & = \frac{1}{2} \mathrm{~g} \mathrm{t}_0^2 - \frac{1}{2} \mathrm{~g} (\mathrm{t}_0^2 - 2 \mathrm{t}_0 \Delta \mathrm{t} + (\Delta \mathrm{t})^2) \\ & = \frac{1}{2} \mathrm{~g} \cdot (2 \mathrm{t}_0 \Delta \mathrm{t} - (\Delta \mathrm{t})^2) \\ & = \mathrm{g} \cdot (\mathrm{t}_0 \Delta \mathrm{t} - \frac{1}{2} \Delta \mathrm{t}^2) \end{aligned} $

Expression for $ \mathrm{t}_0 $:

Solving the above equation for $ \mathrm{t}_0 $, we have:

$ \mathrm{H} = \mathrm{g} (\mathrm{t}_0 \Delta \mathrm{t} - \frac{1}{2} \Delta \mathrm{t}^2) $

$ \mathrm{t}_0 \Delta \mathrm{t} = \frac{\mathrm{H}}{\mathrm{g}} + \frac{1}{2} \Delta \mathrm{t}^2 $

$ \mathrm{t}_0 = \frac{\mathrm{H}}{\mathrm{g} \Delta \mathrm{t}} + \frac{\Delta \mathrm{t}}{2} $

Therefore, the time $ \mathrm{t}_0 $ at which the separation between the two stones becomes $ \mathrm{H} $ is given by:

$ \mathrm{t}_0 = \frac{\mathrm{H}}{\mathrm{g} \Delta \mathrm{t}} + \frac{\Delta \mathrm{t}}{2} $

Motion In A Straight Line

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