The ratio of the displacements of a freely fa | vedclass

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(D) The displacement of a body in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.For a freely falling body,the initial v

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$3: 9$

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(D) The displacement of a body in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.For a freely falling body,the initial velocity $u = 0$ and acceleration $a = g$.Thus,the displacement in the $n^{th}$ second is $S_n = \frac{g}{2}(2n - 1)$.For the second second $(n = 2)$: $S_2 = \frac{g}{2}(2(2) - 1) = \frac{g}{2}(3) = 1.5g$.For the fifth second $(n = 5)$: $S_5 = \frac{g}{2}(2(5) - 1) = \frac{g}{2}(9) = 4.5g$.The ratio of displacements is $\frac{S_2}{S_5} = \frac{1.5g}{4.5g} = \frac{1.5}{4.5} = \frac{1}{3}$.

The ratio of the displacements of a freely falling body during the second and fifth seconds of its motion is

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