Bernoulli equation (watering can)

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SUMMARY

This discussion focuses on applying Bernoulli's equation to a practical problem involving a watering can. The gardener needs to determine the horizontal distance (x) the water must travel when the mouth of the can is at a height (h) of 0.8m, given a constant water level (H) of 1m. The solution involves calculating the velocity of the water using Bernoulli's equation and deriving the maximum distance (x) based on the height (h). The critical height for maximum distance is found to be 0.5m, yielding a maximum distance of 1m.

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JulienB
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Homework Statement



Hi everybody! Here is a classical Bernoulli problem, which I'd like you to review to check if I (finally!) make a proper use of Bernoulli's equation!

A gardener waters a bush with the help of a watering can (see attached pic). The water level in the can lays at height H = 1m, and the water flows horizontally out of the can at height h.
a) at which distance x from the bush must the water flow when the mouth of the can is located at h = 0.8m, in order for the water to reach the bush?
b) for which height h is x maximum? (we consider H is constant)

Homework Equations



Bernoulli equation, free fall and maybe a little bit of derivative?

The Attempt at a Solution



I think I get the point of the problem, but I often get confused :biggrin: Hopefully this time I get it right:

a) I first set up a Bernoulli equation between points (0) (behavior at height H) and (1) (mouth of the can at height h):

ρg(H - h) = ½⋅ρ⋅v12
(Here I considered P0 = P1 = 0 at gauge pressure, v0 = 0 and h the reference height)
⇔ v1 = √(2⋅g⋅(H - h)) = √(0.4⋅g)

Now from basic equations of motion I know:

horizontal displacement: x = v1⋅t
vertical displacement: -h = -½⋅g⋅t2

I solve for t using the second equation and get:

t = √(2h/g)

...and substitute it in the first one:

x = v1⋅t = √(0.8⋅h) = 0.8m

Is that correct?

b) Here I used the equation I found for velocity in a):

x = v⋅t = √(2g⋅(H - h))⋅√(2h/g) = 2√(hH - h2)

Here I wasn't sure what I should do so I simply took the derivative of (hH - h2) with respect to h and solved for which h it is equal to 0 (that is, when the function reaches a local critical point):

f'(h) = (hH - h2)' = H - 2h
f'(h) = 0 ⇔ h = ½⋅H = 0.5 m

Does that make sense? I get x = 1m, which is at least indeed bigger than in a). I'm sure there is also an easier method to get that result, but I couldn't figure it out yet. Any idea?Thank you very much in advance for your answers, I appreciate it.Julien.
 

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All looks good, well done.
 
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Likes   Reactions: JulienB
@haruspex thank you, that is definitely thanks to your huge help! :) I hope this post can help other people too!

Julien.
 

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