From the top of a tower, a particle is thrown vertically downwards with a velocity of $10ms^{- 1}$. The ratio of the distances covered by it in the 3rd and 2nd seconds of its motion is (Given, $g = 10ms^{- 2}$)

Question

Dr. Abubacker · Accepted Answer

(a) $S_{3rd} = 10 + \frac{10}{2}(2 	imes 3 - 1) = 35$
$S_{2nd} = 10 + \frac{10}{2}(2 	imes 2 - 1) = 25$
$	herefore\ \frac{S_{3rd}}{S_{2nd}} = \frac{35}{25}\ ie = \frac{7}{5}$

Q. From the top of a tower, a particle is thrown vertically downwards with a velocity of $10ms^{- 1}$. The ratio of the distances covered by it in the 3rd and 2nd seconds of its motion is (Given, $g = 10ms^{- 2}$)

Explanation

Reviewed by Dr. Abubacker

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