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A particle is thrown up inside a stationary lift of sufficient height and the time of flight is T. Now it is thrown again with same initial speed V0 with respect to the lift. At the time of second throw, lift is moving up with speed V0 and uniform acceleration g (where g is the the acceleration due to gravity). The new time of flight will be.


A
T4
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B
T2
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C
T
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D
2T
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Solution

The correct option is C T2
For the first throw, time of flight T=2V0g
For the second throw, the velocity of particle is V0 w.r.t lift.
Vrel=V0
Displacement of particle upwards would be same as displacement of lift upwards, so relative displacement has to be zero. i.e Srel=0
Relative acceleration of particle w.r.t lift arel=gg=2g
Using second equation of motion,
Srel=urelt+arelt22
0=V0t2gt22
Solving this we get t=0 and t=V0g
t=0 is not possible
t=V0g=T2

Physics

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