Water drops fall at regular intervals from a tap the tap $5 m$ above the ground. The third drop is leaving the tap, the instant the first drop touches the ground. How far above the ground is the second drop at that instant? $(g = 10 {ms}^{- 2})$

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$4.00 m$

$3.75 m$

$2.50 m$

$1.25 m$

answer is B.

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Detailed Solution

$t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 5}{10}} = 1 s$

Water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap, the instant the first drop touches the ground. How far above

Let $t_{0}$ is the interval between two drops. Then
$2 t_{0} = t$
$t_{0} = 0.5 s$
$2^{nd}$ drop has taken $t_{0}$ time to fall. Therefore distance fallen,

$d = \frac{1}{2} g t_{0}^{2} = (\frac{1}{2}) (10) (0.5)^{2} = 1.25 m$

Height from ground = $h - d$
$= 5 - 1.25 = 3.75 m$

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