(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

An inverted garbage can of weight W is suspended in air by water from a geyser. The water shoots up at a constant rate [itex]\frac{\mathrm{d} m}{\mathrm{d} t}[/itex] with velocity [itex]v_{0}[/itex]. What is the maximum height at which the garbage can rides? The upwards direction is positive. Answer clue: if [itex]v_{0} = 20\frac{m}{s}, W = 98.1N, \frac{\mathrm{d} m}{\mathrm{d} t} = .5\frac{kg}{s}[/itex] then [itex]h_{max} \approx 17m[/itex].

3. The attempt at a solution

The net force on the garbage can is [itex]F = F_{I} - W[/itex] where [itex]F_{I}[/itex] is the impact force due to the stream of water. Consider a small amount of the stream of droplets, [itex]\Delta m[/itex] that, at time [itex]t = t_{0}[/itex], is just about to hit the can. We have that [itex]P_{0} = v_{0}\Delta m[/itex] and at [itex]t_{0} + \Delta t[/itex], the time instantly after the rebound\reflection of the small amount of droplets, [itex]P_{f} = -v_{0}\Delta m[/itex] so [itex]\frac{\Delta P}{\Delta t} = \frac{-v_{0}\Delta m - v_{0}\Delta m}{\Delta t} = \frac{-2v_{0}\Delta m}{\Delta t}[/itex]. Therefore, [itex]\lim_{t \rightarrow 0}\frac{\Delta P}{\Delta t} = \frac{\mathrm{d} P}{\mathrm{d} t} = -2v_{0}\frac{\mathrm{d} m}{\mathrm{d} t} = F_{I}[/itex] thus [itex]\frac{W}{g}a = -(W + 2v_{0}\frac{\mathrm{d} m}{\mathrm{d} t})[/itex] so [itex]a = \frac{-(W + 2v_{0}\frac{\mathrm{d} m}{\mathrm{d} t})}{\frac{W}{g}}[/itex]. At the maximum height, the can is instantaneously at rest so [itex]v_{0}^{2} = -2ah_{max}[/itex]. Then, [itex]h_{max} = \frac{Wv_{0}^{2}}{2g(W + 2v_{0}\frac{\mathrm{d} m}{\mathrm{d} t})}[/itex]. Now if I plug in the parameters from the answer clue then I do indeed get [itex]h_{max} \approx 17m[/itex] (16.93480102m to be exact when using g = 9.81m\s^2). My only problem is I had to use a major assumption when arriving at this solution: the water droplets would be rebounded\reflected perfectly so that the velocity of the small amount of droplets instantly after the rebound would be the same as the velocity instantly before hitting the can. If I drop this assumption then I really can't conceive of any way to find the new velocity using only the given information since I wont know how much of the velocity was lost to the can upon impact. If I keep the assumption, I can't exactly convince myself physically why the assumption is true. Any help would be appreciated, thanks!

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# Homework Help: Inverted garbage can and water hydrant

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