Question

A coin is dropped in a lift. It takes time t₁ to reach the floor when lift is stationary. It takes time t₂ when lift is moving up with constant acceleration. Then

Easy

Solution

For stationary lift $t_{1} = \sqrt{\frac{2 h}{g}}$ and when the lift is moving up with constant acceleration $t_{2} = \sqrt{\frac{2 h}{g + a}}$

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