Diameter of liquid stream at maximum height

[Joe Ramsey]

I'm very lost !

First off thanks to HallsoIvy and AD for answering my last question. However I am still having a difficult time with the last question.
I know that we are suppose to try these problems before we post…but I am having a horrible time understanding how to derive equations. I am at a loss and I do not know how to even approach this problem. Could someone at least try to get me going?

Here is the question:

A nozzle squirts out a stream of liquid of diameter dnozzle. The nozzle is angled at q above the horizontal. Show that the diameter the stream when it is at its maximum height (dstream) is given by the equation at right (in other words, derive the equation), and therefore does not depend on the rate at which the liquid is coming out of the nozzle, the type of liquid, or anything else.

d_stream= d_nozzel/(square root of cos theta)


Does anyone have any good tips on deriving equations?

 

[Doc Al]

Originally posted by Joe Ramsey
A nozzle squirts out a stream of liquid of diameter dnozzle. The nozzle is angled at q above the horizontal. Show that the diameter the stream when it is at its maximum height (dstream) is given by the equation at right (in other words, derive the equation), and therefore does not depend on the rate at which the liquid is coming out of the nozzle, the type of liquid, or anything else.

d_stream= d_nozzel/(square root of cos theta)

Along a stream, how does speed relate to cross section? Consider that at any point in the stream, the amount of water flowing per unit time had better be the same. (Otherwise, where's it going?)

Then figure out what the speed of the water is at maximum height compared to its speed at the nozzle.

 

[Joe Ramsey]

Please unsubscribe me from your service.
Thanks,
Joe Ramsey

 

[JohnDubYa]

I don't think he liked your answer, Doc. :)

 

[Doc Al]

I guess not.