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A liquid-filled tank of height h is placed on a platform also of height h above the ground. A small hole is punched at a vertical distance y from the free surface of the liquid to achieve maximum horizontal range xₘ on the ground. Then which of the following is/are correct?
- xₘ = 2h
- xₘ = 1.5 h
- y = h
- y = 0.75 h
Correct answer: y = 0.75 h
Solution
The jet exits with speed sqrt(2gy) and falls a total height of (2h - y) before hitting the ground. The range x = sqrt(2gy) * sqrt(2(2h-y)/g) = 2*sqrt(y(2h-y)). Maximizing x² = 4y(2h-y): d/dy[y(2h-y)] = 2h - 2y = 0 => y = h. Then xₘ = 2*sqrt(h*h) = 2h.
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