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When an object is thrown upwards, when it eventually comes to rest and starts falling, for how long is it stationary? What about an particle in electric field having an initial velocity towards it's same charge? That too would come to rest and reverse velocity, the question is for how long is it at full stop with rest?

Classical physics gives a time of 0, but is that correct? Is it really at rest for 0 seconds? That answer seems a bit counter intuitive.

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closed as not a real question by David Z Jan 19 '11 at 21:30

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Classical mechanics is the reasonable way to describe something you throw up in the air. By editing your original question, you are basically saying "I don't like the answer I got the first time. I'm going to have you answer the question again and again until I hear what I want to hear." – Mark Eichenlaub Jan 19 '11 at 21:36
@MarkEichenlaub : That conclusion is is baseless, By editing my question all it says that with additional data I have refined the question to be a better and more precise question. I couldn't care less about who is answering the question as long as the answer points to the right question to ask, and be it doesnt even have to answer the better question. Understanding of what is needed to be understood is just as important as making it to be understood or even more. – Arjang Jun 9 '15 at 10:19
up vote 3 down vote accepted

Another result is Earnshaw's_theorem, which tells us that no particle can ever be held in a stable equilibrium in the presense of any other collection of interacting particles.

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That is a very interesting result, but coming to rest is different than being at equilibrium (?)( I think ) – Arjang Jan 19 '11 at 21:20

Assuming you are talking about classical physics and point particles, the answer is zero seconds precisely. Why? Suppose the time were non-zero, so that particle stays at one place for at least time $T$. That means that it's velocity is zero for as much long and also acceleration would be zero and external forces would be zero. But that is a contradiction with the presence of gravity.

The discussion of an extended body is similar because you can repeat everything that was said above for its center of mass. But the object might not rest at all because of its intrinsic rotation/vibration/etc.

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Since classical physics is superseded, in what framework is there an answer other than 0 time? 0 time just seemed wrong! – Arjang Jan 19 '11 at 21:08
In quantum mechanics it doesn't even make sense to talk about particle being located at a precise point or stopping completely. You could possibly obtain positive answer if you created some kind of gravitational attractor. But that would be a different question altogether. – Marek Jan 19 '11 at 21:27
Dear Arjang, in quantum mechanics, the answer is even shorter than the classical answer. ;-) In classical physics, one had a differentiable trajectory, and for an infinitesimal amount of time, it stayed constant near the maximum. However, in quantum mechanics, the typical trajectory contributing to the path integral is non-differentiable almost everywhere, so it is infinitely unlikely that the particle stays at the same place, even for an infinitesimal period of time! However, in QM, particle may stay exactly in the same energy state at a finite probability - because energy is quantized. – Luboš Motl Jan 19 '11 at 21:45
How should I ask this question to find out how long before a moving object/particle reverses it's direction? – Arjang Jan 19 '11 at 21:48
@Arjang: well, formulate the precise conditions you want to consider (what kind of gravitational field, air resistance, etc.) and try your luck. If the question is not well-posed but will have potential for being good someone will help you improve it. – Marek Jan 19 '11 at 21:50

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