Water drops fall at regular intervals from a tap 5 meters above the ground. The third drop leaves the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant?
Give your answer in meters.
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We can calculate the taken by 1st drop to reach the ground as T = g 2 × h = 1 S e c Thus distance travelled by 2nd drop in half second is D = 2 1 × g × 4 1 = 1 . 2 5 M So height of the drop is 5 − 1 . 2 5 = 3 . 7 5 M