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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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5:7

7:5

3:5

5:3

answer is B.

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Detailed Solution

$S_{n} = u + a (n - \frac{1}{2})$
$\frac{S_{3}}{S_{2}} = \frac{10 + 10 (3 - \frac{1}{2})}{10 + 10 (2 - \frac{1}{2})} = \frac{35}{25} = \frac{7}{5}$

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