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If a stone dropped from the top of a building travels half of the distance during its last second it's time period is

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\\\text{Let total distance traveled by the stone} = h$ \\ Now, Velocity at midpoint, $v=\sqrt{2 g s}=\sqrt{g h}$ \\ Velocity at lowest point, $V=v=\sqrt{2 g s}=\sqrt{2 g h}$ \\ Now, $V=v+g t \quad(t=1 s)$ \\ \\ Now, $\sqrt{2 g h}=\sqrt{g h}+g$ \\ \\ $\Rightarrow \sqrt{h}=\frac{\sqrt{g}}{\sqrt{2}-1}$ \\ \\

\\Also, h=\frac{1}{2} g t^{2}$ \\ \\ $\Rightarrow t^{2}-\frac{2 h}{g}=0$ \\ \\ $t=\pm \sqrt{\frac{2 \hbar}{g}}=\pm \frac{\sqrt{2}}{\sqrt{2}-1}$ \\ \\ $t=2 \pm \sqrt{2}$

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