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A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is:

1. π (v⁴)/(g²)
2. (π v⁴)/(2g²)
3. (π v²)/(g²)
4. (π v²)/(g)

Correct answer: π (v⁴)/(g²)

Solution

Water reaches farthest at 45 deg, giving max range R = v^2/g. The wet region is a circle of that radius, so area = pi*R^2 = pi*v^4/g^2.

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