What is the ratio of the distances covered by | vedclass

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(B) For a body falling freely from rest,the initial velocity $u = 0$ and acceleration $a = g$.The distance covered in the $n^{th}$ second is given by

Vedclass · Accepted Answer

$1:3:5$

Vedclass · Answer

(B) For a body falling freely from rest,the initial velocity $u = 0$ and acceleration $a = g$.The distance covered in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.Substituting $u = 0$ and $a = g$,we get $S_n = \frac{g}{2}(2n - 1)$.For $n = 1$: $S_1 = \frac{g}{2}(2(1) - 1) = \frac{g}{2}(1) = \frac{g}{2}$.For $n = 2$: $S_2 = \frac{g}{2}(2(2) - 1) = \frac{g}{2}(3) = \frac{3g}{2}$.For $n = 3$: $S_3 = \frac{g}{2}(2(3) - 1) = \frac{g}{2}(5) = \frac{5g}{2}$.The ratio of distances $S_1 : S_2 : S_3$ is $\frac{g}{2} : \frac{3g}{2} : \frac{5g}{2} = 1 : 3 : 5$.

What is the ratio of the distances covered by a freely falling body in the $1^{st}$,$2^{nd}$,and $3^{rd}$ seconds of its motion?

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