Does a ball thrown in the air really stop at its apex, and if it does, wouldn't that violate the uncertainty principle?

When throwing a ball straight up, most experts say that it momentarily comes to a stop at its apex before its return fall. If it stops, wouldn't we know its velocity and position and wouldn't this violate the uncertainty principle?

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The ball, unfortunately, is made up of about eleventy gajillion quantum particles, each of which are orbiting around one another. – Arkamis Mar 4 '13 at 18:51
Experts only ever state this as an approximation. – Meow Mar 4 '13 at 19:15
Why is the point where it "stops" different from any other point we can say what its velocity and position is according to Newton's laws? It makes no sense to pick on that one point just because the vertical velocity component happens to be 0. – Olin Lathrop Mar 5 '13 at 0:54
@OlinLathrop it is a good question because the turning point is well defined, a "measurement" infinitely accurate in Newtonian physics. – anna v Mar 5 '13 at 4:37
heisenberg uncertainity principle doesn't apply in macroscopic world or we can't observe the effect. – idiosincrasia23 Mar 5 '13 at 11:05

If you model the ball using classical mechanics, then it makes sense to ask whether a ball (under ideal circumstances) stops at its apex when thrown and the answer in that context is yes. However, notice that the question itself depends on treating the ball as an object which travels along a well-defined trajectory in space.

In the context of quantum mechanics, it no longer makes sense to treat the motion of the ball in this way. Instead, the state of the ball at any given time would be modeled as a vector in a Hilbert space, and one could ask questions about ensemble averages of physical quantities such as position and momentum. The Heisenberg uncertainty relation, which is really a bound on standard deviations of sets of measurements made on ensembles of identically prepared systems, would hold.

In particular, if the ball is prepared in a state such that at time $t$, a measurement of its position will yield a value close to the apex of a classical solution, then there will be large uncertainty in the measurement of its momentum at that time.

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The others have answered well for a ball - which consists of many particles so quantum uncertainty is totally unmeasurable. You can however do the experiment with neutrons by bouncing them off special mirrors and you do get discrete energy levels and quantum mechanical "smearing" of the bounce just like you would expect from 1st year quantum mechanics:

(lovingly stolen from the very cool experimental collaboration here)

Notice the spreading of the wavepackets. You can do a back of the envelope calculation using the uncertainty principle and get the right order of magnitude for the spread, which can be found exactly from the Schrodinger equation in this case.

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A simple calculation:

A simple analysis with a couple of examples I hope will reveal the significance of the smallness of $h$ and $\hbar.$

The whole quantum mechanical structure is based on the smallness of Planck’s constant $h=6.625\times 10^{-34}$ Js, or the reduced Planck’s constant $\hbar=1.054\times 10^{-34}$ Js.

Let us assume there is an uncertainty $\Delta x=0.20$m in the position of the ball as it reaches the apex. This would imply an uncertainty in the momentum of the ball given by Heisenberg’s principle as

$\Delta p = \frac{\hbar}{\Delta x}= 5.27\times 10^{-34}$Kgms$^{-1}$.

This momentum of quantum mechanical origin is so minute compared to the classical uncertainty in the momentum of the ball that is impossible to resolve in a measurement. So for all practical purposes, in macroscopic the world, any quantum mechanical effects are insignificant compared to other quantities involved in the problem, as they are impossible to resolve.

A similar analysis holds for the Double slit experiment by using for example grains of rice. Let us assume we have a double slit device with slit distance $d=0.002$m, mass of grains of rice m=0.0005Kg moving at speed $v=100$ms$^{-1}$. The distance of the diaphragm from the screen is $D=2$m, so then we have

$s=\frac{d\lambda}{D}$ where $\lambda = \frac{h}{mv}$ gives $s\sim 10^{-36}$m.

Such small fringe spacing is impossible to have any observable effects on the way grains of rice hit the screen! Impossible to observe any interference pattern by using classical objects such as grains of rice.

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you lost a - sign in front of the value of h_bar in the second line also the 10^-34 in the reduced expression. I will edit – anna v Mar 5 '13 at 4:39
@anna v Many thanks for the edit. It was very late at night when I was composing the answer! – JKL Mar 5 '13 at 10:45