In a lift moving up with an acceleration of 5ms−2, a ball is dropped from a height of 1.25m. The time taken by the ball to reach the floor of the lift is approximately (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 y-axis as positive Relative velocity of the lift with respect to ball ulb=0m/s Distance travelled by the lift w.r.t ball =Slb=1.25m Acceleration of the lift =a=5m/s2 Acceleration due to gravity =g=−10m/s2 (because a and g are in opposite direction) Acceleration of the lift w.r.t ball =alb=a–(−g)=5–(−10)=15m/s2 Using second equation of motion, we get Slb=ulb.t+12albt2 ⇒1.25=12×15×t2 t=√2×1.25m(5+10)ms−2≈0.4s Hence, the ball will take nearly 0.4 seconds to touch the floor of the lift.