The ratio of distance covered by a body in $1 S, 2 S, 3 S, ....... n s$ is

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$1 : 4 : 9 : ..... n^{2}$

$1 : 2 : 3 : ...... n$

$1 : \frac{1}{2} : \frac{1}{3} : ....... \frac{1}{n}$

$1 : 1 : 1 : ....$

answer is A.

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Detailed Solution

$e q u a t i o n o f m o t i o n, S = U t + \frac{1}{2} g t^{2} i n i t i a l v e l o c i t y = U = 0; a c c e l e r a t i o n d u e t o g r a v i t y = g; t i m e e l a p s e d = t; d i s p l a c e m e n t = S S = h e i g h t = \frac{1}{2} g t^{2}$

In 1st sec the distance traveled is, (put t=1 then) $S_{1} = \frac{1}{2} g t^{2} = \frac{g}{2}$
In 2nd sec the distance traveled is,( put t=2 then) $S_{2} = \frac{1}{2} g t^{2} = \frac{4 g}{2}$
In 3rd sec the distance traveled is( put t=3 then) $S_{3} = \frac{1}{2} g t^{2} = \frac{9 g}{2}$
$S_{1} : S_{2} : S_{3} : ....... = \frac{g}{2} : \frac{4 g}{2} : \frac{9 g}{2} : .......... S_{1} : S_{2} : S_{3} : ....... = 1 : 4 : 9 : 16$