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

Question

Dr. Abubacker · Accepted Answer

(b) $S_{3^{rd}} = 10 + \frac{10}{2}(2 	imes 3 - 1) = 35\ m$
$S_{2^{nd}} = 10 + \frac{10}{2}(2 	imes 2 - 1) = 25\ m \Rightarrow \frac{S_{3^{rd}}}{S_{2^{nd}}} = \frac{7}{5}$

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

Explanation

Reviewed by Dr. Abubacker

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