The speed of light upon reflection (NOTE: I am an 8th grader, so I may not be capable of perfectly elaborating my point by scientific measures.  Also, English is not my first language)
I have read that the speed of light is constant, and does not decrease unless under certain circumstances (such as going through dense and/or thick material.)
When a ball hits the wall and rebounds, it has momentum, except for the moment that it hits the wall, since the speed would be 0 there.
How about light?  How can light maintain a constant speed upon hitting something?  Shouldn't the speed of the photons be zero when they are neither moving forward nor backward when they have hit the wall?
 A: In this case, it would be useful to not consider light in its particle form as photons, but instead to consider it as a wave - see this wikipedia page. Then, the wave is simply reflected from the surface, without us having to consider the kinetics of any particle. The wave, in a vacuum, would continue to propagate at the speed of light, regardless of the surface it is reflected off of.
However, it is possible to carry on considering the light as photons. These photos are absorbed by the material, which then instantaneously emits a new photon with the same velocity as the incident photon. This doesn't involve the photon being reflected by the material, so avoids the case in which a photon would have a velocity of 0.
Thanks to TheGhostOfPerdition for pointing out that I had missed that second part.
A: Let me explain in purely classical terms (not the description of reality, but easy to imagine).
You realized that when a ball bounces off the wall, at a certain point, it has no momentum. However, it must still have all the energy of the movement (neglecting losses to environment) - where did that energy go?
The ball is formed of many discrete atoms, bound together by electromagnetic forces. As the ball starts hitting the wall, it compresses - the atoms are pushed closer together than they would without the pressure. As the back of the ball approaches the front of the ball, the kinetic energy is converted into that compression, and the momentum is transferred to the wall (and through it, the Earth, accelerating its rotation by a miniscule amount). At some point, the kinetic energy is all gone - into the compressed ball, the sound, slight heating of both the ball and the wall etc. But the ball is still compressed, and the atoms don't want to be so close together - so they push on each other, and on the atoms in the wall. The stored compression energy is converted back into kinetic energy, and momentum is "stolen back" from the wall (again, accelerating the Earth's rotation by a tiny amount). The ball picks up speed, and when all the compression is released, it travels at the same speed as when it hit (ignoring energy losses to the environment).
Photon isn't a composite particle. There's nothing to "compress" to store the kinetic energy. Compare how a rubber ball bounces to the way a billiard ball bounces. In the first case, a lot of the energy is stored in the rubber ball's compression. But a billiard ball is a lot harder, and almost no energy is stored in its compression - the time the ball spends in the "no momentum" stage is much shorter than for the rubber ball. You can consider a photon to be infinitely hard in this picture - it doesn't spend any time at all in the "compressed" phase - it just bounces right off, just like balls in a perfect newton's craddle.
However, I have to note again that this is not how light actually really works. In reality, the photon doesn't really bounce off at all - rather, it is absorbed, and after a bit of time, another photon is emitted in the right direction, as if the photon bounced off.
And the nice thing about physics is that as you go deeper, you'll see a lot of "well, that's not quite the case". All of those behaviours you were taught (and will be taught in the future) are in our models of reality, not reality itself. They are useful models for modelling certain scenarios, and useless for others. Need to calculate where a beam of light will hit if you shine it on a mirror? You can already do that, just with the simple Brewster's angle model. But reality is more complex than that - for example, if you grind off pieces of the mirror at precisely calculated places, you'll have a mirror that reflects light at a different angle (look up "diffraction mirror" if this sounds interesting :)).
To the best of our knowledge, even photons aren't really real. Right now, the best model we have of reality has no balls bouncing around, no photons sometimes behaving like little balls and sometimes like waves in a pool, there's just a quantum field that is neither a little ball nor a wave in a pool. But we only use that model when it's absolutely necessary - because it's very difficult to use. But we know that's closest to reality than anything else we have, because it explains (and predicts) things that no other theory does, and we already depend on the theory in real practical devices in everyday use (I assume you have a personal computer, since you're posting questions on StackExchange :P). Someday, a better theory might gain ground, and we'll be saying "Well, we thought light is really a surface phenomenon of an underlying electromagnetic quantum field, but it so happens that was a silly thing to think! Instead, bananas are really the fundamental thing in the universe, obviously." :)
