Water drops are falling at a regular interval of time from a certain height H above the ground. At the instant the fourth drop is released, the first one just reaches the ground. The height of the third drop from the ground is

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$\frac{H}{4}$

$\frac{3 H}{4}$

$\frac{8 H}{9}$

$\frac{H}{9}$

answer is C.

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Detailed Solution

Let the time interval be t then $H = \frac{1}{2} g {(3 t)}^{2} \Rightarrow \frac{1}{2} g {(t)}^{2} = \frac{H}{9}$

The height of 3^rd drop from the ground is = $H - \frac{1}{2} g {(t)}^{2} = H - \frac{H}{9} = \frac{8 H}{9}$

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