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 ofsecond 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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a
2T
b
T2
c
T
d
2T
answer is B.
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Detailed Solution
With respect to lift initial speed = v0Acceleration =- 2g Displacement =0∴S=ut+12at2 0=v0T′+12×2g×T′2∴ T′=v0g=12×2v0g=12T