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12 Aug 2012 21:50:23 IST

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One dimension motion

Engineering Entrance , Medical Entrance , AIPMT , JEE Main , AIIMS , JEE Advanced , Physics

A stone falls freely from rest and the total distance covered by it in the last second of its motion equals to the distance covered by it in first three seconds of its motion. The stone remains in air for how many seconds

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Vikram Saxena

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Joined: 1 Apr 2007

Posts: 4025

13 Aug 2012 22:22:25 IST

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The equation to be used is distance formula : s=ut+.5*g*t^2. So distance in first 3 sec = .5*g*3*3, distance in last second=.5*g*t^2- .5*g*(t-1)^2 where t is the total time.equating both the equations we get, t=5secs.

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