Galileos law of odd numbers:
"The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, $1 : 3 : 5 : 7..$ ]". Prove it.

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Solution

$S_{0 t o 1} = \frac{1}{2} a t^{2} = \frac{1}{2} a$
$S_{1 t o 2} = \frac{1}{2} a (2)^{2} - \frac{1}{2} a (1)^{2} = \frac{1}{2} a (3)$
$S_{2 t o 3} = \frac{1}{2} a (3)^{2} - \frac{1}{2} a (2)^{2} = \frac{1}{2} a (5)$
$S_{n - 1 t o n} = \frac{1}{2} a (n)^{2} - \frac{1}{2} a (n - 1)^{2} = \frac{1}{2} a (2 n - 1)$
Therefore, the ratio is $1 : 3 : 5 : 7 : . . . . . . . . . . . : (2 n - 3) : (2 n - 1)$

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Similar questions

1 : 3 : 5 : 7........

The distances traversed by a freely falling body, falling from rest, in equal intervals of time is in the following ratio ____________.

1^{st}, 2^{nd}, 3^{rd}

g = 10 {m/s}^{2}

A

a_{1}

3

B

a_{2}

6^{th} second

A

a_{1} : a_{2}

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