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A tank with vertical walls sits with its base at height H above level ground. Water fills it to depth h. A hole is made in the side wall at depth x below the water surface. For the emerging jet to have maximum horizontal range on the ground, what is x?

1. (H + h)/4
2. (H + h)/2
3. (H + h)/3
4. 3(H + h)/4

Correct answer: (H + h)/2

Solution

The jet leaves horizontally at v = sqrt(2gx) and falls a height (H + h - x) to the ground. Range R = 2*sqrt(x(H + h - x)). The product x(H + h - x) is maximized when x = (H + h)/2.

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