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(B) Let the distance $AB = BC = h$. The total distance $AC = 2h$.Since the body falls freely from rest at $A$, the initial velocity $u = 0$.Using the

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$\sqrt{2} - 1$

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(B) Let the distance $AB = BC = h$. The total distance $AC = 2h$.Since the body falls freely from rest at $A$, the initial velocity $u = 0$.Using the equation of motion $s = \frac{1}{2}gt^2$, the time $t_1$ taken to travel distance $AB = h$ is:$t_1 = \sqrt{\frac{2h}{g}}$The total time $T$ taken to travel distance $AC = 2h$ is:$T = \sqrt{\frac{2(2h)}{g}} = \sqrt{\frac{4h}{g}} = 2\sqrt{\frac{h}{g}}$The time $t_2$ taken to travel distance $BC$ is the difference between the total time $T$ and the time $t_1$:$t_2 = T - t_1 = 2\sqrt{\frac{h}{g}} - \sqrt{\frac{2h}{g}} = \sqrt{\frac{h}{g}}(2 - \sqrt{2})$Now, the ratio $(t_2 / t_1)$ is:$\frac{t_2}{t_1} = \frac{\sqrt{\frac{h}{g}}(2 - \sqrt{2})}{\sqrt{\frac{2h}{g}}} = \frac{2 - \sqrt{2}}{\sqrt{2}} = \frac{2}{\sqrt{2}} - \frac{\sqrt{2}}{\sqrt{2}} = \sqrt{2} - 1$

$A, B, C$ are points in a vertical line such that $AB = BC$. If a body falls freely from rest at $A$, and $t_1$ and $t_2$ are the times taken to travel distances $AB$ and $BC$ respectively, then the ratio $(t_2 / t_1)$ is:

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