A window is 50,cm long. A stone is falling fr | vedclass

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(C) Let the top of the window be at distance $h_1 = 40\,cm$ from the starting point of the stone.Let the bottom of the window be at distance $h_2 = 40

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$\frac{1}{7}\,s$

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(C) Let the top of the window be at distance $h_1 = 40\,cm$ from the starting point of the stone.Let the bottom of the window be at distance $h_2 = 40\,cm + 50\,cm = 90\,cm$ from the starting point.The time taken to reach the top of the window is $t_1 = \sqrt{\frac{2h_1}{g}} = \sqrt{\frac{2 	imes 40}{980}} = \sqrt{\frac{80}{980}} = \sqrt{\frac{4}{49}} = \frac{2}{7}\,s$.The time taken to reach the bottom of the window is $t_2 = \sqrt{\frac{2h_2}{g}} = \sqrt{\frac{2 	imes 90}{980}} = \sqrt{\frac{180}{980}} = \sqrt{\frac{9}{49}} = \frac{3}{7}\,s$.The time taken to cross the window is $\Delta t = t_2 - t_1 = \frac{3}{7} - \frac{2}{7} = \frac{1}{7}\,s$.

$A$ window is $50\,cm$ long. $A$ stone is falling from a height of $40\,cm$ above the window. It crosses the window in

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