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

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

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

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