From the top of a tower, a particle is thrown vertically downwards with a velocity of 10 m/s. The

From the top of a tower, a particle is thrown vertically downwards with a velocity of 10 m/s. The ratio of the distances, covered by it in the 3rd and 2nd seconds of the motion is (Take g=10 m/s 2)

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[sanatsingh]

Great sir I searched for solutions everywhere but you provided the right expalination