A lift accelerates vertically downward at 4 m s⁻². A particle is projected upward inside it with speed 10 m s⁻¹ relative to the lift. Taking g=10 m s⁻², the time taken to return to the point of projection in the lift is:
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Question

A lift accelerates vertically downward at 4 m s⁻². A particle is projected upward inside it with speed 10 m s⁻¹ relative to the lift. Taking g=10 m s⁻², the time taken to return to the point of projection in the lift is:

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G1 Physics · Accepted Answer

Relative acceleration of the particle with respect to the lift is downward with magnitude effective gravity = g−4 = 6 m s⁻². Therefore the return time is T = 2u/effective gravity = 20/6 = 10/3 s.