The ratio of the distances travelled by a fre | vedclass

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Question

(B) The distance travelled by a body in the $n^{	ext{th}}$ second is given by the formula: $S_{n} = u + \frac{a}{2}(2n - 1)$.Since the body is freely

Vedclass · Accepted Answer

$1: 3: 5: 7$

Vedclass · Answer

(B) The distance travelled by a body in the $n^{	ext{th}}$ second is given by the formula: $S_{n} = u + \frac{a}{2}(2n - 1)$.Since the body is freely falling,the initial velocity $u = 0$ and acceleration $a = g$.Thus,$S_{n} = \frac{g}{2}(2n - 1)$.This implies that $S_{n} \propto (2n - 1)$.For the $1^{	ext{st}}, 2^{	ext{nd}}, 3^{	ext{rd}}$,and $4^{	ext{th}}$ seconds,the ratio is:$S_{1} : S_{2} : S_{3} : S_{4} = (2(1) - 1) : (2(2) - 1) : (2(3) - 1) : (2(4) - 1)$.$S_{1} : S_{2} : S_{3} : S_{4} = 1 : 3 : 5 : 7$.

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

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