A person is standing in a stationery lift drops a coin from a certain height h.It takes time t to reach the floor of the lift. If the lift is rising up with uniform acceleration a , the time taken by the coin, dropped from the same height h , to reach the floor will be ________

Dear student,

Please find below the solution to your asked query.

initial velocity u = 0
when lift is at rest,
h = ut+0.5gt²
h = 0+0.5gt²
when lift rises up with some constant acceleration a , then
effective acceleration a' = g+a
so
h = 0+0.5(g+a)t'²
therefore
0.5(g+a)t'² = 0.5gt²
$$\begin{gathered} t \text{'} = \left(\sqrt{\frac{g}{g+a}}\right)t \\ t\text{'} =t {\left(1+\frac{a}{g}\right)}^{-1/2} \end{gathered}$$
where t' is the time taken by the coin to reach to the bottom when lift is accelerated in upward direction.

Hope this information will doubt about laws of motion.

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