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Question

In a lift moving up with an acceleration of 5 ms−2, a ball is dropped from a height of 1.25 m. The time taken by the ball to reach the floor of the lift is approximately (g=10 ms−2)

A
0.3 second
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B
0.2 second
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C
0.16 second
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D
0.4 second
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Solution

The correct option is D 0.4 second
Taking upward y-axis as positive
Relative velocity of the lift with respect to ball ulb=0 m/s
Distance travelled by the lift w.r.t ball =Slb=1.25 m
Acceleration of the lift =a=5 m/s2
Acceleration due to gravity =g=10 m/s2 (because a and g are in opposite direction)
Acceleration of the lift w.r.t ball =alb=a(g)=5(10)=15 m/s2
Using second equation of motion, we get
Slb=ulb.t+12albt2
1.25=12×15×t2
t=2×1.25 m(5+10) ms20.4 s
Hence, the ball will take nearly 0.4 seconds to touch the floor of the lift.

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