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I'm studying Torricelli's law and in Wikipedia the diagram below says that the height of the intersection of two streams of water is equal to the sum of their source depths.

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Torricelli's law describes the parting speed of a jet of water, based on the distance below the surface at which the jet starts, assuming no air resistance, viscosity, or other hindrance to the fluid flow. This diagram shows several such jets, vertically aligned, leaving the reservoir horizontally. In this case, the jets have an envelope (a concept also due to Torricelli) which is a line descending at 45 degrees from the water's surface over the jets. Each jet reaches farther than any other jet at the point where it touches the envelope, which is at twice the depth of the jet's source. The depth at which two jets cross is the sum of their source depths. Every jet (even if not leaving horizontally) takes a parabolic path whose directrix is the surface of the water.

How is this derived? I'm currently trying to prove this using concepts from projectile motion and I kept falling on a deadlock $y_1=y_2$. I already know that $v = \sqrt{2gh}$. I'm currently assuming that the initial velocities at the x-axis are $\sqrt{2gh_1}$ and $\sqrt{2gh_2}$ while the initial velocities at the y-axis are both zero (where $h_1$ and $h_2$ are the source depths).

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  • $\begingroup$ Hi Wilhelm, I have edited your post , but if there is anything you want to change, just click the edit button below the tags $\endgroup$ – user108787 Nov 12 '16 at 14:37
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From there, it is quite straightforward.

Imagine that at t=0, water particles leave from both the openings. Now, it is obvious that the time taken by them to reach the intersection will not be same because their horizontal speeds (which are unchanged after leaving the opening) are unequal.

So, lets say particle from the top opening takes $t_{1}$ time to reach the intersection and particle from the lower one takes $t_{2}$ time. All we have to do is to equate the distances traveled along x axis and y axis.

Equating distance along x axis, $$\sqrt{2gh_{1}}t_{1}=\sqrt{2gh_{2}} t_{2}$$ Equating distance from top along y axis, $$h_{1}+\frac{1}{2}gt_{1}^2=h_{2}+\frac{1}{2}gt_{2}^2$$

Solving those two, we get $$t_{2}=\sqrt{\frac{2h_{1}}{g}}$$ Substituting in the distance of molecule from the lower opening from top, $$depth=h_{2}+\frac{1}{2}g\frac{2h_{1}}{g}$$ $$\Longrightarrow depth=h_{1}+h_{2}$$ On a side note, we can also get x distance as $2\sqrt{h_{1}h_{2}}$. So we have the location of the intersection.

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