A lift starts moving up with an acceleration of 5ms−2, and at the same instant a ball is dropped from a height of 1.25m. The time taken by the ball to reach the floor of the lift is (nearly) (g=10ms−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 D0.4 second Taking upward as positive,
Initial velocity of the ball =vb=0
Initial velocity of the lift ul=0
Distance travelled by the lift w.r.t ball is given by Slb=1.25m
Acceleration of the lift =a=5m/s2
Acceleration of the ball =g=−10m/s2 (because a and g are in opposite direction)
Using second equation of motion, we can say Slb=ulbt+12albt2 ⇒1.25=0+12(5+10)t2 ⇒t=√2×1.255+10 ≈0.4s
Hence, the ball will take 0.4 seconds to touch the floor of the lift.