Fluid Dynamics: Solving the Equation of Continuity

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Homework Help Overview

The problem involves fluid dynamics, specifically the equation of continuity in the context of a hose watering a lawn. The scenario describes a situation where blocking part of the hose opening affects the distance the water reaches, raising questions about the relationship between velocity and area in fluid flow.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of the equation A1v1=A2v2 and question the relationship between velocity and distance in projectile motion. There are attempts to clarify the meaning of time consistency in the context of the problem.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem setup, particularly regarding the angle of the hose and the initial height of the water. Some guidance has been offered about neglecting initial height and considering the angle of projection.

Contextual Notes

There is uncertainty regarding the initial height of the hose and the angle at which the water is projected, which affects the interpretation of the problem. Participants are encouraged to consider these factors as they analyze the situation.

joseph_kijewski
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Homework Statement



You are watering your lawn with a hose when you put your finger over the hose opening to increase the distance the water reaches. If you are pointing the hose at the same angle, and the distance the water reaches increases by a factor of 4, what fraction of the hose opening did you block?

Homework Equations

The Attempt at a Solution



This seems very obvious to me. The density of the water would remain consistent, thus the equation would become A1v1=A2v2. Since t is consistent as long as the height the water is fired from is, I figured this means v2 must be 4 times v1. But the answer is 2 times? I've looked over this problem many times and I just can't see what I'm missing, it seems so obvious?
 
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joseph_kijewski said:
Since t is consistent as long as the height the water is fired from is, I figured this means v2 must be 4 times v1.
What do you mean by "t is consistent"? You'll need to review projectile motion to see how the range of a projectile depends on initial speed.
 
TSny said:
What do you mean by "t is consistent"? You'll need to review projectile motion to see how the range of a projectile depends on initial speed.

Shouldn't x component be v*t, thus directly proportional to velocity?? By t is consistent, I mean that t must be consistent in both as h is consistent: -4.9t^2=h
 
When a football is kicked into the air, does the time of flight change if the initial speed is changed?

4.9t2 = h is for horizontal projection from a height h.

The statement of the problem is not clear about whether or not to take into account the initial height of the hose. I suspect that you are meant to neglect the initial height and imagine that the water essentially leaves the hose at ground level.
 
When you write the equation x = vt, what does v represent? Is it the initial speed, or is it the x-component of the initial velocity, or is it the y-component of the initial velocity, or something else?
 
I guess I'm imagining the hose as horizontal, in which case I believe I would be right, but the angle isn't specified, so isn't it impossible to determine the answer?
 
joseph_kijewski said:
I guess I'm imagining the hose as horizontal, in which case I believe I would be right, but the angle isn't specified, so isn't it impossible to determine the answer?
Yes, if you interpret the problem as aiming the hose horizontally from some height h, then to quadruple the range you would need to quadruple the initial speed. As you noted, this is not the answer they wanted.

The problem statement mentions "pointing the hose at the same angle". This suggests that the hose is not horizontal, but tilted upward. However, the problem is still not clear regarding the initial height. See if you get the "right" answer if you assume the initial height is small enough to neglect. So, the water essentially leaves the hose at ground level at some unknown angle θo.
 
Figured it out, thanks!
 
OK. Good work!
 

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