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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/s²):

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3 : 6

7 : 5

6 : 3

5 : 7

answer is B.

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Detailed Solution

$\begin{array}{l} S_{3 rd} = 10 + \frac{10}{2} (2 \times 3 - 1) = 35 m \\ S_{2 nd} = 10 + \frac{10}{2} (2 \times 2 - 1) = 25 m \Rightarrow \frac{S_{3 rd}}{S_{2 nd}} = \frac{7}{5} \end{array}$

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