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

\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 A $\frac{T}{2}$
time of flight for first throw is given by $T = \frac{2 v}{g}$
but in second case $v$ is same but relative acceleration of ball wrt to the lift $2 g$ i.e. $(g + g)$ . so time of flight will reduce to half that of first one.

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Q. A particle is thrown up with initial speed

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and uniform acceleration

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The correct option is A T2time of flight for first throw is given by T=2vgbut in second case v is same but relative acceleration of ball wrt to the lift 2g i.e.(g+g). so time of flight will reduce to half that of first one.

The correct option is A $\frac{T}{2}$
time of flight for first throw is given by $T = \frac{2 v}{g}$
but in second case $v$ is same but relative acceleration of ball wrt to the lift $2 g$ i.e. $(g + g)$ . so time of flight will reduce to half that of first one.