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Question

# A very broad elevator is going up vertically with a constant acceleration of 2 m/s2. At the instant when its velocity is 4 m/s a ball is projected from the floor of the lift with a speed of 4 m/s relative to the floor at an elevation of 30∘. The time taken by the ball to return the floor is (g = 10 m/s2) from time of projection.

A

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B

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C

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D

1 s

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Solution

## The correct option is B aballlift=aball−alift =−g^j−2^j =−(g+2)^j =−12ms2^j Now in the lifts frame, the ball is simply a projectile with a = −12ms2^j and not a = =−gms2^j ∴ Time of flight = T = 2μ sin θ12 = 2×4×sin3012 = 2 × 4 × 12 × 112 = 13s

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