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Water reach (Hose without and with a nozzle)

  • #1

Homework Statement


A hose of inside diameter ##1.5cm## can reach a distance of ##1.5m##. A nozzle is inserted in the hose and water can now reach ##24m##. What is the inside diameter of the nozzle? The height is the same in both cases.

Homework Equations


Use continuity equation ##v_{hose}A_{hose}=v_{nozzle}A_{nozzle}##

The Attempt at a Solution


Continuity equation gives ##1.5/t\pi r_{hose}^{2}=24\pi r_{nozzle}^{2}## which is the same as ##1.5(0.015/2)^{2}=24r_{nozzle}^{2}##. So ##r_{nozzle}=0.1875cm## and the inside diameter will be twice that or ##d=0.375cm##.

The solution however is ##d=0.75cm## (twice what I got). Where am I wrong?
 
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Answers and Replies

  • #2
haruspex
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What is the relationship between launch speed and range?
 
  • #3
What is the relationship between launch speed and range?
##x=v_{horizontal}t=v_{horizonta}\sqrt{\frac{2h}{g}}##. ##h## is the same in both cases
 
  • #4
haruspex
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h is the same in both cases
I see no reason why either t or h should be the same in both. What should we assume is the same?
 
  • #5
I see no reason why either t or h should be the same in both. What should we assume is the same?
The problem states the height is the same in both cases (with and without the nozzle)
 
  • #6
haruspex
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The problem states the height is the same in both cases (with and without the nozzle)
I think they mean the heights of the end of the hose and of the point reached by the jet are the same.
 
  • #7
I think they mean the heights of the end of the hose and of the point reached by the jet are the same.
Yes exactly
 
  • #8
haruspex
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Yes exactly
But you have used it in post #3 as though the height the water reaches at the top of its trajectory is the same in both scenarios. That is not the case.
 
  • #9
But you have used it in post #3 as though the height the water reaches at the top of its trajectory is the same in both scenarios. That is not the case.
I see. Still if the water exits with only horizontal velocity it can only fall down not go up. That means the point the water reaches has to be at a lower height.
 
  • #10
haruspex
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if the water exits with only horizontal velocity
It says "can reach". What does that imply about the angle of the hose?
 
  • #11
It says "can reach". What does that imply about the angle of the hose?
If the angle is 90 with the horizontal obviously cant reach. So some angle between 0 and 90 degrees. I am imagining it must be 45 for some reason but I dont see what reason
 
  • #12
gneill
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If the angle is 90 with the horizontal obviously cant reach. So some angle between 0 and 90 degrees. I am imagining it must be 45 for some reason but I dont see what reason
Look up the "range equation" for projectile motion.
 
  • #13
haruspex
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must be 45 for some reason but I dont see what reason
Because that gives the maximum range. But you do not need to assume it is 45 to solve the problem; you only have to assume that the angle does not change.
 

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