- #1
faust9
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Ok, I'm vexed... I was given the following problem:
The water is shown leaving the nozzle along a 3-4-5 triangle where the horizontal component is 4 and the verticle component is 3 (or 36.1'ish degrees).
So, my question is how do I find velocity given an angle and a curvature? I know radius of curvature ([itex]\rho[/itex]) is:
[tex] \rho=\frac{[1+(y^{\prime})^2]^{2/3}}{y^{\prime \prime}}[/tex]
I have no idea where to go from here. My professor told us the answer was 16.57m/s but I haven't the foggiest clue on how to get there. Any help would be greatly appreciated.
Thanks.
From measurements of a photograph it has been found that the stream of water leaving a nozzle at A had a radius of curvature of 35m.
(a) determine the initial velocity
(b) determine the radius of curvature at hmax
The water is shown leaving the nozzle along a 3-4-5 triangle where the horizontal component is 4 and the verticle component is 3 (or 36.1'ish degrees).
So, my question is how do I find velocity given an angle and a curvature? I know radius of curvature ([itex]\rho[/itex]) is:
[tex] \rho=\frac{[1+(y^{\prime})^2]^{2/3}}{y^{\prime \prime}}[/tex]
I have no idea where to go from here. My professor told us the answer was 16.57m/s but I haven't the foggiest clue on how to get there. Any help would be greatly appreciated.
Thanks.