Discussion
A tank open at the top with a water level of 1 m, as shown in the figure, has a hole at a height of 0.5 m. A free jet leaves horizontally from the smooth hole. The distance X (in m) where the jet strikes the floor is
Answer: B
The jet strikes the floor 1.0 m from the tank.
The hole is 0.5 m below the water surface, so by Torricelli's theorem the jet leaves horizontally at \(v = \sqrt{2g \times 0.5} = \sqrt{g}\). It then falls the 0.5 m from the hole to the floor:
\[ \begin{aligned} t &= \sqrt{\frac{2 \times 0.5}{g}} = \frac{1}{\sqrt{g}} \\ X &= vt = \sqrt{g} \times \frac{1}{\sqrt{g}} = 1.0\ \text{m} \end{aligned} \]
In general X = \(2\sqrt{h_{1}h_{2}}\), with \(h_{1}\) the depth of the hole below the surface and \(h_{2}\) its height above the floor; here \(2\sqrt{0.5 \times 0.5} = 1\). 0.5 m halves this, and 2.0 or 4.0 m would need a much greater head or drop.
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