I play volleyball once a week in a local rec center. There is a relatively low ceiling and some other odds and ends that the ball can deflect against. As I understand it, in a normal bump the ball should ascend in a parabola, and should return to a player at approximately the same speed as it was hit upwards.
When the ball hits the ceiling before completing a full arc, it seems to return faster than if it just goes up and falls back down unimpeded. Is this an optical illusion or is the return speed somehow increased because the ball has not achieved apogee before it rebounds?
In a normal arc, all the upward inertial force is finally exhausted by the time it gets to the apogee. In a ceiling hit, most of the remaining energy is redirected downward. Thus you have the normal acceleration due to gravity, plus the velocity imparted to the ball by the elastic energy of the bounce.
Someone will be along shortly to translate that into proper physics, and correct my carelessness with force, energy, etc.
Mmmmmmm, don’t think this one requires any physics lingo or detailed explanation… the ball bounces down off the ceiling, so it would naturally come down faster than if you had just dropped it. Just get one of those little rubber bouncy balls - hold it up near the ceiling and let it fall. Then throw the ball up at the ceiling and watch how much faster it “falls”. Fairly simple.
If you neglect air resistance and assume an elastic collision (these two simplifications are often made in Physics problems), then the ball should return at the same speed. So, I guess I have to disagree with you, dqa. Someone may be along shortly to correct me, though…
But let me explain exactly what I mean. Suppose you hit the ball up in the air, it reaches 14 feet, and falls back and hits you in the face, travelling at the same speed as when you first hit it. Now suppose you put a 12-foot ceiling in, and hit the ball at the same speed upward. It will again hit you in the face at the same speed that you hit it. However, if your ceiling is of uniform height, then of course balls that bounce off the ceiling will return faster than ones that don’t, because if the ceiling weren’t there, the ones that are bouncing are the ones that would be going the highest. Does that make sense?
Now, about those assumptions I made. The one about the elastic collision is probably pretty sound, I imagine. The one about air resistance is probably not so good. If you do take air resistance into account, then in my previous example, the one that bounces off the 12-foot ceiling would return faster than the one which reached 14 feet, because it spends less time in the air. The less time it spends, the less energy it loses to friction. I’m not sure how much faster it would return, though. I guess, then, if you were taking air resistance into account, dqa, I agree with you.
Oh, finally, I want to point out that the example I used involved a ball being shot straight up, but if you served it at an arc, everything I said would still apply.
That’s an oversimplification, or a misunderstanding of my question. I’ll give an example to clarify:
I hit a ball that will normall go 40 feet in the air, then return down. If the ceiling is a 25 feet, does the ball come down faster than it would have if had already descended (and accelerated) for fifteen feet from the point of apogee.
So it’s not the same thing as dropping the ball from the ceiling, it would be the same as dropping the ball from fifteen feet above the ceiling. I think the first answer sounds right, it’s kind of what I was thinking but couldn’t really express. The key is that instead of travelling a total of 80 feet, it would now travel a distance of only 50 feet. So if both balls were hit with the same force, the one which is deflected back by the ceiling would have to arrive back sooner since it has a shorter path to cover.
Let’s assume no air resistance and perfectly elastic collisions. Then if you throw a ball at a certain speed, it will come down at exactly the same speed you threw it at - it will lose some speed going up, and gain the same a mount of speed coming down. If it hits the ceiling or some obstacle on the way, it will still come down with exactly the same speed. Think about it in terms of energy - energy of the ball is kinetic plus potential. Potential energy before and after the flight is the same, because it is at the same height (give or take a couple of feet). Elastic collisions don’t rob energy. So kinetic energy before and after the flight must be the same as well - hence the speed must be the same.
Of course, you have to throw the ball hard if you want it to hit the ceiling. So any ball that hits the ceiling will come down faster than a ball that didn’t make it to the ceiling.
If you include air resistance, the difference becomes even larger. If a ball hits the ceiling it comes down quicker, so it spends less time travelling in the air - and has less time to lose speed due to air resistance.
“I hit a ball that will normall go 40 feet in the air, then return down. If the ceiling is a 25 feet, does the ball come down faster than it would have if had already descended (and accelerated) for fifteen feet from the point of apogee.”
No. At least, not to first order. If you use the simplifying assumptions which scr4 and I mentioned, it will return at exactly the same speed. However, if air resistance is significant, it will tend to make the ceiling ball return faster. If the collision is inelastic, it will tend to make the ceiling ball return more slowly. I’m not sure which effect will be greater in real life, or even if either one is significant.
Both of these complications can be thought of in terms of energy, since the speed that the ball returns with is related to the energy that it has. In the case of air resistance, objects lose energy by traveling through air, so the one that goes further will have less energy. In the case of inelastic collisions, objects lose energy by hitting other things, so the one that bounces will lose energy.
Well I was assuming that the ball would be hitting the cieling lower. Like comparing it from throwing it 40 ft in the air as opposed to hitting a roof 25 ft high with the same initial energy.
