A body starts to fall freely under gravity. T | vedclass

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(A) The distance covered by a body in the $n^{th}$ second is given by the formula: $S_n = u + \frac{g}{2}(2n - 1)$.Since the body starts from rest,the

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$1:3:5$

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(A) The distance covered by a body in the $n^{th}$ second is given by the formula: $S_n = u + \frac{g}{2}(2n - 1)$.Since the body starts from rest,the initial velocity $u = 0$.Therefore,the distance covered in the $n^{th}$ second is $S_n = \frac{g}{2}(2n - 1)$.For the first second $(n=1)$: $S_1 = \frac{g}{2}(2(1) - 1) = \frac{g}{2}(1) = \frac{g}{2}$.For the second second $(n=2)$: $S_2 = \frac{g}{2}(2(2) - 1) = \frac{g}{2}(3) = \frac{3g}{2}$.For the third second $(n=3)$: $S_3 = \frac{g}{2}(2(3) - 1) = \frac{g}{2}(5) = \frac{5g}{2}$.Taking the ratio $S_1 : S_2 : S_3$,we get $\frac{g}{2} : \frac{3g}{2} : \frac{5g}{2} = 1 : 3 : 5$.

$A$ body starts to fall freely under gravity. The distances covered by it in the first,second,and third seconds are in the ratio:

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