A body is dropped from a height H. The time t | vedclass

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(C) Let $T$ be the total time taken to fall from height $H$. Using the equation of motion $s = ut + \frac{1}{2}gt^2$ with initial velocity $u = 0$:$H

Vedclass · Accepted Answer

$\sqrt{\frac{H}{g}}(\sqrt{2}-1)$

Vedclass · Answer

(C) Let $T$ be the total time taken to fall from height $H$. Using the equation of motion $s = ut + \frac{1}{2}gt^2$ with initial velocity $u = 0$:$H = \frac{1}{2} g T^2 \Rightarrow T = \sqrt{\frac{2H}{g}}$Let $t_1$ be the time taken to cover the first half of the journey,i.e.,distance $H/2$:$\frac{H}{2} = \frac{1}{2} g t_1^2 \Rightarrow t_1 = \sqrt{\frac{H}{g}}$The time taken to cover the second half of the journey is the difference between the total time and the time taken for the first half:$t_2 = T - t_1 = \sqrt{\frac{2H}{g}} - \sqrt{\frac{H}{g}}$$t_2 = \sqrt{\frac{H}{g}} (\sqrt{2} - 1)$

$A$ body is dropped from a height $H$. The time taken to cover the second half of the journey is:

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