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Let A, B, C, D be points on a vertical line such that $AB = BC = CD$. If a body is released from position A, the times of descent through AB, BC and CD are in the ratio.
Explanation
$S = AB = \frac{1}{2} gt_{1}^{2} \Rightarrow 2 S = AC = \frac{1}{2} g(t_{1} + t_{2})^{2}$ and $3S = AD = \frac{1}{2}g(t_{1} + t_{2} + t_{3})^{2}$ $t_{1} = \sqrt{2S/g}$ $t_{1} + t_{2} = \sqrt{4S/g}$, $t_{2} = \sqrt{4S/g} - \sqrt{2S/g}$ $t_{1} + t_{2} + t_{3} = \sqrt{6S/g}$ $t_{3} = \sqrt{6S/g} - \sqrt{4S/g}$ $t_{1}:t_{2}:t_{3}$:: $1:(\sqrt{2} - \sqrt{1}):(\sqrt{3} - \sqrt{2})$