A particle is thrown up inside a stationary lift of sufficient height. The time off light is T. Now it is thrown again with same initial speed v₀ with respect to lift. At the time of
second throw, lift is moving up with speed v₀ and uniform acceleration g upward (the acceleration due to gravity). The new time of flight is

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$\sqrt{2} T$

$\frac{T}{2}$

$T$

$2 T$

answer is B.

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Detailed Solution

With respect to lift initial speed = v₀
Acceleration =- 2g

$\begin{array}{l} Displacement = 0 ∴ S = u t + \frac{1}{2} a t^{2} \\ \begin{array}{l} 0 = v_{0} T^{'} + \frac{1}{2} \times 2 g \times T^{' 2} \\ ∴ T^{'} = \frac{v_{0}}{g} = \frac{1}{2} \times \frac{2 v_{0}}{g} = \frac{1}{2} T \end{array} \end{array}$