Kleppner and Kolenkow, 2nd edition, problem 4.23 - Suspended garbage can I am working through Kleppner and Kolenkow's An Introduction to Mechanics on my own and have a question about the solution of the mentioned problem.
Problem Statement: An inverted garbage can of weight $W$ is suspended in air by water from a geyser.  The water shoots up from the ground with a speed of $v_0$ at a constant rate $K$ kg/s.  The problem is to find the maximum height at which the garbage can rides.  Neglect the effect of the water falling away from the garbage can.
The book/TA solution I have found is quite nice and uses $\bf{F}_{tot} = \dot{\bf{P}}_{in}$ from the text. It also uses a fully elastic collision of the water and bucket, so that the momentum transfer and force are doubled.
My question is how to work this problem using $P(t)$ and $P(t+\Delta t)$, as is done in sections 4.7 and 4.8 of the text.
Here is what I have, which doesn't work.  I think I probably have setup the problem incorrectly:
$P(t) = Mv + \Delta m u$
$P(t+\Delta t) = (M + \Delta m)(v+\Delta v)$
which gives $\frac{dP}{dt} = M \frac{dv}{dt} + (v-u)\frac{dm}{dt} = -Mg = F_{tot}$
and $u = v_0 - gt$.
Substituting in $u$, then solving the first order ODE and eventually eliminating $t$ leads to
$h = \frac{1}{2g}(\frac{2Mg}{K} + v_0)^2$, which is incorrect.
I have also tried $P(t+\Delta t) = M(v+\Delta v)  -\Delta m u$, to account for the elastic collision, but this leads to a 3rd power after integration that does not work out either.
Any help would be appreciated, thanks.
 A: 
UPDATE:
After more attempts on this problem, I have come up with a solution that yields the answer from the solutions manual (not the one in the back of the book, which is wrong, but from the actual solution manual), using the setup my original question asked for.  The only part I'm slightly uneasy about is when I set $du/dt = 0$, but I think it is ok.  I'm also not currently sure how this setup would be extended to include the effect of water falling away from the garbage can, but that was not part of the original question.
Any comments on the solution would be appreciated.  Thanks!
My Solution
Let $M$ be the mass of some blob of water from the jet, let $v$ be the velocity of the blob and let $u$ be the velocity of the bucket.  Consider a moment before the blob hits the bucket, and the moment after:
$P(t) = M v + M_b u$
$P(t+\Delta t) = -Mv + M_b (u + \Delta u)$
Then,
$\Delta P = -2Mv + \Delta u M_b$
Note that $M = K\Delta t$, and if we divide the above by $\Delta t$ and take the limit as $\Delta t \rightarrow 0$, then,
$\frac{dP}{dt} = -2 v K + \frac{du}{dt} M_b = - M_b g$
or
$\frac{2 v k}{M_b} - \frac{du}{dt} = g$
Now, when the can reaches its maximum height, $u=0$ and hence $du/dt = 0$, which substituted in yields,
$\frac{2 v^* k}{M_b} = g$
or
$v^* = \frac{M_b g}{2K} = \frac{W}{2K}$.
From applying conservation of momentum to the blob, and eliminating $t$, we also have $v = \sqrt{v_0^2 - 2gh}$,  where $h$ is the height of the blob.  Equating these last two equations and solving for $h^*$, where we have used $v^*$, yields
$h^* = \frac{1}{2g}\left(v_0^2 - \left(\frac{W}{2K}\right)^2 \right)$
which is the solution provided in the solution manual.
A: The Two equations you started with in the question, seems incorrect to me! Have a Look at my approach.

