1. The problem statement, all variables and given/known data 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) 2. Relevant equations Bernoulli equation, free fall and maybe a little bit of derivative? 3. The attempt at a solution I think I get the point of the problem, but I often get confused 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.