Bernoulli equation (watering can)

AI Thread Summary
The discussion revolves around a Bernoulli equation problem involving a gardener using a watering can to water a bush. The key points include calculating the horizontal distance (x) the water must travel when the can's mouth is at a height of 0.8m, and determining the height (h) for maximum distance (x). The user derives the velocity of water flow using Bernoulli's equation and applies kinematic equations to find the horizontal displacement. The calculations suggest that the maximum distance occurs when the height is 0.5m, resulting in a distance of 1m. The user expresses gratitude for assistance received and hopes the discussion aids others.
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 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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