SMBC ball bouncing problem This comes from a Saturday Morning Breakfast Cereal (SMBC) comic with a joke answer. The problem states:

A 5 kilogram ball is shot directly right at 20 meters per second from a height of 10 meters. The ball loses 1 joule whenever it touches Earth. Assume no air resistance. When does the ball stop bouncing?

How would one solve this problem? The best I could do was to assume the total energy of the ball, given by the sum of potential and kinetic energy when it's initially shot is completely lost when it stops bouncing. This would give us approximately 1490 bounces, with each bounce slowing the ball down and making it bounce ever so slightly lower.
This still requires a ton of calculation (a huge series), even with the added assumption that there is no friction between the ball and the ground. Am I missing something?
 A: If one assumes that momentum is lost in both directions, since kinetic energy is nondirectional, perhaps it is best to assume that it loses energy in each "direction" proportionally to the sine and cosine of the angle of the bounce. Using this assumption, each bounce causes it to shed 2/3 J from horizontal velocity and 1/3 from vertical potential energy. Under this assumption, there are 1491 bounces, each linearly smaller than the last, coming out finally to roughly 250km in distance.
A: 
When does the ball stop bouncing?

Using kinematics means the height ($d_y$) goes to zero and vertical velocity ($v_y$) goes to zero, but the horizontal velocity ($v_x$) would be constant if friction and air resistance are negligible, since it would continue to slide after bouncing has stopped. So you could solve for when $d_y$ = 0 or $v_y$ = 0. 
Using conservation of energy, potential energy ($PE_i$) equals the kinetic energy ($KE_i$), minus 1.0 J lost in each inelastic collision. Since there is no force in the horizontal direction (negligible friction and air resistance), then I will assume the energy dissipated in the bounces is only based on the energy in the vertical direction, which could be the work due to gravity (assuming no energy is lost due to deformation of the material). Using the vertical mechanical energy ($E_y$) as the initial potential energy gives us:
$$E_y = PE_i = mgh \\ = (5)(9.81)(10) \\ = 490.5\ \ J$$
When the ball stops bouncing, the vertical mechanical energy ($E_y$) equals the sum of the energy lost in each bounce ($E_{bounce}$), let n be the number of bounces:
$$E_y = \sum E_{bounce} \\ 490.5\ J = (1.0\ J)n \\ n = 490.5\ \rm{bounces} $$
So the ball stops bouncing after 490.5 bounces, which is really after 490 bounces. (I don't have time to solve for t).
EDIT, July 15th:
Gravity would likely not be dissipative (just as gravity provides a restoring force in a pendulum). Therefore the energy must be due to deformation in the material, which would not allow it to be broken down into x and y components as shown above.
So, the total mechanical energy should be calculated as the sum of the potential and kinetic energies and set equal to the sum of the energy dissipated in the bounces.
$$E_T = mgh + \frac{1}{2}mv^2 = \sum E_{bounce} \\ \\ (5)(9.81)(10) + \frac{1}{2}(5)(20)^2 = (1.0\ J)n \\ \\ n = 1490.5\ \rm{bounces} $$
Assuming no friction or resistance, and rounding up, the ball would bounce 1491 times before coming to rest.
A: The comments to @udiboy's answer point out that this problem is a bit ill posed (which, for a comic strip, is ok I guess). There's seems to be some argument above about whether friction can be neglected (I note that the question does NOT say to neglect friction) and whether the gravitational potential energy can be depleted without touching the horizontal component of momentum.
Realistically, the ball probably loses a bit of speed and height with every bounce, and the process is impossible to calculate without more information about the ground-ball interaction. But we can still get a constraint on the time.
I see two limiting cases here:
1) The horizontal momentum is unaffected by bounces (maybe the ball is frictionless but "sticky"?). The ball bounces to a slightly lower height after each bounce, and finishes sliding horizontally along the ground with the same horizontal velocity it had initially until the end times. udiboy solved this in his answer, so I'll shamelessly steal his result, and call it the minimum time until bouncing stops:
$$T_\mathrm{min} = \sqrt{\frac{h_i}{g}} + \sum_1^n \sqrt{\frac{2}{g}\frac{E_i-n}{mg}}$$
2) The ball loses horizontal momentum with every bounce, but bounces to the same height every time, until it is out of horizontal momentum. Then it loses height with each bounce, until it is out of height, and finishes at rest. This isn't very realistic, but it is an upper bound on the time:
$$T_\mathrm{max} = T_\mathrm{min} + \sqrt{\frac{h_i}{2g}}mv_i^2$$
Taking $g$ to be $10\textrm{ms}^{-2}$ gives (provided I haven't botched computing the sum):
$$T_\mathrm{min} = 472\mathrm{s}$$
$$T_\mathrm{max} = 1886\mathrm{s}$$
Ok, so I didn't say it was going to be a good constraint. But it's better than I'm used to getting (ah, the joys of astronomy).
A: If we assume no friction, there is no force in the horizontal direction so the ball will continue to move indefinately towards the right. We can consider it to stop bouncing when it has no vertical velocity. As the forces are completely in the vertical direction, there is no energy loss from the kinetic energy in the horizontal direction. Thus that 1 Joule dissipated will be from the potential energy. PE is given by 
$$\Delta P=mgh=500 J$$.
So we can say it will bounce five hundred times.
After the first bounce, the PE will become $\Delta P=499 J$ and the height corresponding to this PE will be
$$\Delta P=mgh_1$$ $$\therefore h_1=\frac{\Delta P}{mg}=\frac {499}{50}m$$
This means that after the first bounce the ball goes up to a height of $\frac{499}{50}m$. Similarly, after the $n^{th}$ bounce
$$h_n=\frac {500-n}{50}$$
The time taken for a ball to go to a height of $h$ and come back under the force of gravity is given by
$$t=\sqrt{\frac{2h}{g}}$$
Thus time taken from the $n^{th}$ bounce to the ${n+1}^{th}$ bounce is 
$$t_n=\sqrt{\frac{2h_n}{g}}$$
So, a summation of this should give you the answer.
$$T_{total}=\sum_1^n t_n$$
Note that we have started summing from after the first bounce. We will also have to add the initial time it takes to fall, which is given by $$t_0=\sqrt{\frac{h_{initial}}{g}}$$
The problem with considering friction: Now there is a force in the horizontal direction, so it will do work each time the ball bounces. To find the time, we will have to find out exactly what part of the $1 Joule$ is being dissipated by friction and inelastic collision. This becomes really complex. I don't know how to do it.
A: Since we don't know the volume of the ball, we may approximate it as a point particle. Some mention has been made of the friction between the ground and the ball - all I will say is that we should ignore it because a) the moment of inertia (or some way to find it) aren't specified, and b) if the ball did start spinning, the friction would become static, not kinetic. Still a doable problem, but you'd need the moment of inertia.
$$
\begin{eqnarray}
U_0 &=& \frac{m \dot x_0^2}{2} + m g h = 1490\text J\\
U_n &=& \frac{m \dot x_0^2}{2} + m g h - n
\end{eqnarray}$$
Where $m$ is the ball's mass, $v_0$ is the initial (horizontal) velocity, $g$ is the local surface gravity (assumed to be $9.8 \frac{\text m}{ \text s}$), $h$ is the initial height, and $n$ is the number of bounces and has units of Joules. Obviously, we must devise a sum over the energy from 1489 to 1 and add the duration of the first half parabola.
Even though the collision isn't perfectly elastic, I can think of no preference for a shallower or steeper angle after the bounce. Incidentally, that angle is $35^\circ$, and upon reflection, the ratio $\frac{\dot y_n}{\dot x_n} = \frac{\sqrt{2 g h}}{\dot x_0} = \eta \approx \frac{7}{10}$ would then be constant.
The duration of the $n$th bounce is $t = \frac{2 \dot y_n}{g}$ (from kinematics), so all we must do is put $y_n$ in terms of $U_n$. When the ball hits the ground, it only has kinetic energy.
$$
\begin{eqnarray}
U_n &=& \frac{m(\dot x_n^2 + \dot y_n^2)}{2} \\
U_n &=& \frac{m(\frac{\dot y_n^2}{\eta^2} + \dot y_n^2)}{2} \\
U_n &=& \frac{m\dot y_n^2}{2}\left( 1 + \frac{1}{\eta^2}\right) \\
\dot y_n &=& \sqrt{\frac{2 U_n}{m \left(1 + \frac{1}{\eta^2} \right)}} 
\end{eqnarray}
$$
Doing a little algebra and writing the sum, 
$$t_{total} = \sqrt{\frac{2 h}{g}} +  \frac{2 \eta}{g} \sqrt{\frac{2}{m\left(1 + \eta^2\right)}} \sum_{n=1489}^1 \sqrt{U_n} \approx 1471 \text s$$
If you were writing a test, the sum of square roots is a formidable problem, but you can avoid that (using the apocryphal Gauss trick) if you ask for the distance traveled. I leave it as a (quite fun) exercise for the reader.
