Water drops fall at regular intervals from a tap which is S_m above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant
Text Solution
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Time taken by first drop to reach the ground \(\ell = \frac{1}{g} \sqrt{2gh}\)
\(\to t \sqrt{\frac{2 \times 5}{10}}\) 1 sec
As the water drops fall at regular intervals from a tap therefore time difference between any two drops \(-\frac{1}{2} x x\)
In this given time, distance of second drop from the tap \(= \frac{1}{2} g \left| \frac{1}{2} \right|^2 = \frac{5}{8} = 1.25 m\)
Its distance from the ground \(5 \cdot 1.25 \quad 3.75 \, m\)
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