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The water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap at an instant when the first drop touches the ground. How far above the ground is the second drop at that instant? (Take g = 10 m/s^2)

1. (a) 1.25 m
2. (b) 2.50 m
3. (c) 3.75 m
4. (d) 5.00 m

Correct answer: (b) 2.50 m

Solution

The time for the first drop to fall 5 m is calculated using the equation of motion: h = (1/2)gt², giving t = 1 s. Since drops fall at regular intervals, the second drop has been falling for 0.5 s. Using the same equation, its height above the ground is h = 5 - (1/2)(10)(0.5²) = 2.5 m.

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