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Question

# An elevator is moving up with an acceleration a m/s2. A person inside the lift throws the ball upwards with a velocity u m/s relative to himself. Then the time of flight of ball (in sec) will be:

A
ug+a
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B
uga
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C
2ug+a
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D
2uga
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Solution

## The correct option is C 2ug+a The scenario for motion of ball is shown w.r.t lift frame. with respect to lift frame: initial velocity of ball =ur=u m/s acceleration of ball w.r.t lift: ar=−(g+a) m/s2 , −ve sign implies downward direction. ∴ applying the equation of motion in lift frame: Sr=urt+12art2 ...(i) where Sr is the displacement in the lift frame. For t=T=Time of flight, putting Sr=0 in equation (i), we get: 0=uT−12(g+a)T2 ∴ T=2ug+a seconds

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