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JEE PHYSICS MCQ: MOTION IN A STRAIGHT LINE

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From the top of tower, a particle is thrown vertically downwards with a velocity of 10 m/s10 \mathrm{~m} / \mathrm{s}. The ratio of the distances covered by it in the 3rd and 2nd seconds of the motion is : (Take g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 ) :
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Correct option: B
0.295138889
Option B (07:05) is correct because it is the ratio of distances calculated using the formula for distance covered in the nth second.

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