The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second is:

Question

Accepted Answer

1.  **Identify the Law:** The distance traversed by a body falling from rest during equal intervals of time follows **Galileo's Law of Odd Numbers** .
2.  **Mathematical Relation:** The distance ($S_n$) travelled in the $n$-th second of uniformly accelerated motion (starting from rest, $u=0$) is given by:
    $$S_n = \frac{g}{2}(2n - 1)$$
    This shows that $S_n \propto (2n - 1)$.
3.  **Calculate Ratios:**
    *   For $n=1$ (1st second): $S_1 \propto (2(1) - 1) = 1$
    *   For $n=2$ (2nd second): $S_2 \propto (2(2) - 1) = 3$
    *   For $n=3$ (3rd second): $S_3 \propto (2(3) - 1) = 5$
    *   For $n=4$ (4th second): $S_4 \propto (2(4) - 1) = 7$
4.  **Conclusion:** The ratio of distances is $1:3:5:7$.

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