The ratio of times taken by a freely falling | vedclass

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(B) For a body falling freely from rest, the distance $s$ covered in time $t$ is given by $s = \frac{1}{2}gt^2$. Thus, the time taken to cover a dista

Vedclass · Accepted Answer

$1: \sqrt{2}-1: \sqrt{3}-\sqrt{2}$

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(B) For a body falling freely from rest, the distance $s$ covered in time $t$ is given by $s = \frac{1}{2}gt^2$. Thus, the time taken to cover a distance $s$ is $t = \sqrt{\frac{2s}{g}}$. Let $t_1$ be the time to travel the first $5 \ m$ $(s_1 = 5 \ m)$. Then $t_1 = \sqrt{\frac{2(5)}{g}} = \sqrt{\frac{10}{g}}$. Let $t_2$ be the time to travel the first $10 \ m$ $(s_2 = 10 \ m)$. Then $t_2 = \sqrt{\frac{2(10)}{g}} = \sqrt{\frac{20}{g}}$. Let $t_3$ be the time to travel the first $15 \ m$ $(s_3 = 15 \ m)$. Then $t_3 = \sqrt{\frac{2(15)}{g}} = \sqrt{\frac{30}{g}}$. The time taken to travel the first $5 \ m$ is $T_1 = t_1 = \sqrt{\frac{10}{g}}$. The time taken to travel the second $5 \ m$ is $T_2 = t_2 - t_1 = \sqrt{\frac{20}{g}} - \sqrt{\frac{10}{g}} = \sqrt{\frac{10}{g}}(\sqrt{2} - 1)$. The time taken to travel the third $5 \ m$ is $T_3 = t_3 - t_2 = \sqrt{\frac{30}{g}} - \sqrt{\frac{20}{g}} = \sqrt{\frac{10}{g}}(\sqrt{3} - \sqrt{2})$. The ratio $T_1 : T_2 : T_3$ is $\sqrt{\frac{10}{g}} : \sqrt{\frac{10}{g}}(\sqrt{2} - 1) : \sqrt{\frac{10}{g}}(\sqrt{3} - \sqrt{2})$. Simplifying, we get $1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2})$.

The ratio of times taken by a freely falling body to travel the first $5 \ m$, second $5 \ m$, and third $5 \ m$ distances is:

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