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A tap is fixed 5 m above the ground and releases water droplets at equal time intervals. At the moment when the first droplet reaches the ground, the third droplet has just started to leave the tap. At that same instant, what is the height of the second droplet above the ground? (Take g = 10 m/s²)

1. 1.25 m
2. 2.50 m
3. 3.75 m
4. 5.00 m

Correct answer: 3.75 m

Solution

Drop 1 takes t where 5 = (1/2)(10)t^2 -> t = 1 s. Three drops at equal 0.5 s intervals, so drop 2 has fallen 0.5 s: distance = (1/2)(10)(0.5)^2 = 1.25 m, height = 5 - 1.25 = 3.75 m.

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