The ratio of the distance travelled by a freely falling body in the 1^st, 2^nd, 3^rd and 4^th second is: - Physics MCQ NEET 2022 2026 | ExamTest.live

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The ratio of the distance travelled by a freely falling body in the 1^{s t}, 2^{n d}, 3^{r d} and 4^{t h} second is:

Solution & Step-by-step Explanation

The distance travelled by a body in the n^{t h} second (S_{n}) starting from rest (u = 0) is given by: S_{n} = u + \frac{a}{2} (2 n - 1) = \frac{g}{2} (2 n - 1) For n = 1, 2, 3, 4 : S_{1} = \frac{g}{2} (2 (1) - 1) = \frac{g}{2} (1) S_{2} = \frac{g}{2} (2 (2) - 1) = \frac{g}{2} (3) S_{3} = \frac{g}{2} (2 (3) - 1) = \frac{g}{2} (5) S_{4} = \frac{g}{2} (2 (4) - 1) = \frac{g}{2} (7) Ratio S_{1} : S_{2} : S_{3} : S_{4} = 1 : 3 : 5 : 7 .

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The ratio of the distance travelled by a freely falling body in the 1^{s t}, 2^{n d}, 3^{r d} and 4^{t h} second is:

1 : 2 : 3 : 4

1 : 4 : 9 : 16

1 : 3 : 5 : 7

1 : 1 : 1 : 1

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