A coin is dropped in a lift. It takes time ${t_1}$ to reach the floor when lift is stationary. It takes time ${t_2}$ when lift is moving up with constant acceleration. Then
A coin is dropped in a lift. It takes time ${t_1}$ to reach the floor when lift is stationary. It takes time ${t_2}$ when lift is moving up with constant acceleration. Then
${t_1} > {t_2}$
${t_2} > {t_1}$
${t_1} = {t_2}$
${t_1} > > {t_2}$
A coin is dropped in a lift. It takes time ${t_1}$ to reach the floor when lift is stationary. It takes time ${t_2}$ when lift is moving up with constant acceleration. Then
For stationary lift ${t_1} = \sqrt {\frac{{2h}}{g}} $
and when the lift is moving up with constant acceleration
${t_2} = \sqrt {\frac{{2h}}{{g + a}}} $
$\therefore \;\;{t_1} > {t_2}$
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