Question

A particle is thrown up with initial speed V0 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 upwards (where g is the acceleration due to gravity). The new time of flight will be

- T4
- 2T
- T2
- T

Solution

The correct option is **C** T2

Taking upward as positive for both cases,

For the first throw, time of flight T=2V0g

For the second throw, the velocity of particle is V0 with respect to 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 is given by

arel=−g−g=−2g

Using second equation of motion,

Srel=urelt+12arelt2

⇒0=V0t+12arelt2⇒t=V0g=T2

Taking upward as positive for both cases,

For the first throw, time of flight T=2V0g

For the second throw, the velocity of particle is V0 with respect to 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 is given by

arel=−g−g=−2g

Using second equation of motion,

Srel=urelt+12arelt2

⇒0=V0t+12arelt2⇒t=V0g=T2

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