A particle is thrown up with initial speed $V_{0}$ inside a stationary lift of sufficient height and the time of flight is $T$ . Now it is thrown again with same initial speed $V_{0}$ with respect to the lift. At the time of second throw, lift is moving up with speed $V_{0}$ and uniform acceleration $g$ upwards (where $g$ is the acceleration due to gravity). The new time of flight will be

\frac{T}{4}

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\frac{T}{2}

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T

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2 T

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Solution

The correct option is B $\frac{T}{2}$
Taking upward as positive for both cases,
For the first throw, time of flight $T = \frac{2 V_{0}}{g}$
For the second throw, the velocity of particle is $V_{0}$ with respect to lift.
$V_{r e l} = V_{0}$
Displacement of particle upwards would be same as displacement of lift upwards, so relative displacement has to be zero. i.e. $S_{r e l} = 0$
Relative acceleration of particle w.r.t lift is given by
$a_{r e l} = - g - g = - 2 g$
Using second equation of motion,
$S_{r e l} = u_{r e l} t + \frac{1}{2} a_{r e l} t^{2}$
$\Rightarrow 0 = V_{0} t + \frac{1}{2} a_{r e l} t^{2} \Rightarrow t = \frac{V_{0}}{g} = \frac{T}{2}$

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