[Yahya (OP)]

dt can never be pure zero. That is the purpose of "d"

an object at stationary , it elapses zero distance and the time it takes zero at an instant ( because when start to count time will first be zero ) how do you determine its velocity , there is not limits here , zero distance and zero time ? similar problem for the stone. but we all know this velocity is zero , and the stone velocity has the instantaneous velocity , how 0/0=0 ? 0/0=anynumber
but fortunately we know its velocity is zero , because it will still not make displacement and then v=0/t=0 , for the stone its velocity is the instantaneous velocity taken from the slope of the graph.

[jeffreyH]

v dt = dx.

[chris]

In calculus, v is normally expressed as "the limit as dt approaches 0. "
For a well-behaved function (like an object lobbed in a gravitational field), v is well-behaved as dt approaches 0, and when dt = 0. See L'Hopital's rule.

Who was L'Hopital?

[jeffreyH]

In calculus, v is normally expressed as "the limit as dt approaches 0. "
For a well-behaved function (like an object lobbed in a gravitational field), v is well-behaved as dt approaches 0, and when dt = 0. See L'Hopital's rule.

Who was L'Hopital?

https://en.m.wikipedia.org/wiki/Guillaume_de_l'H%C3%B4pital

[Yahya (OP)]

The object will stay exactly as newtonian mechanics would predict- an infinitesimal amount of time.

does that mean that I'm the only one who has this idea?

[Yahya (OP)]

You need to learn calculus to understand the mathematical concepts.

I learned calculus....

v=dx/dt    But dx and dt are not 0.

but I learned that if dx/dt=0/0 that does not mean an answer does not exist.

[evan_au]

  'It depends on whether the flow of time is smooth and indivisible.  Or whether time is made of discrete particles such as "chronotrons" '
If time is quantized, it hasn't shown up at the very small timescale  events probed by the high energies in the LHC.

A Stone has much lower energy per nucleon than particles in the LHC, so we can effectively treat time as continuous.

It is possible that time is quantised on very small timescales; and it may even have macroscopic side-effects. Just as quantised energy levels in atoms avoided the immediate collapse of all atoms in the universe, it is possible that quantised time may someday explain some other phenomenon like persistent infinities around black holes.

But for stones falling in Earth's gravity, treat time as continuous.

[nilak]

The object will stay exactly as newtonian mechanics would predict- an infinitesimal amount of time.

does that mean that I'm the only one who has this idea?

The object really stops at the top and v reaches absolute zero according to newtonian mechanics. However according to my concept this never happens because my concept uses waves propagation to explain motion and all propagations are at a constant speed c and obviously never stop.

[Colin2B]

does that mean that I'm the only one who has this idea?

No. Newton and his peers knew about this which is one of the reasons they were interested in the development of calculus.
The answers are within this thread see

I think the answer is : it spend at that point zero time

This is the traditional answer that Newton would have given.
- The acceleration of the stone is (almost) constant.
- the velocity starts off negative (upwards) and increases in an (almost) straight line until it reaches the same value, but positive (downwards)
- the amount of time spent crossing the v=0 axis is 0 seconds
- this answer doesn't really change if you add in the inverse square law of gravity, or air resistance

but the confusion can be better explained if you avoid talking about an object only having a vertical component of velocity.
Consider an object thrown with both vertical and horizontal velocity components. The object will reach the apex and have zero velocity at a specific time - the turnaround point. That time is a specific point on the curve and if you zoom in to that point it gets smaller and smaller trending towards 0 in both velocity and time. So the problem is not another Zeno paradox as suggested earlier, but a fractal effect.
If we look at the way Newtonian mechanics calculates the total time the object is in flight we find it is symmetric around the apex ie no time is allowed for any hold or hesitation at the apex - this you will find on many school homework sites.
As @evan_au says, it can only be considered to spend any time at the apex if time is quantised and as yet we have no evidence for that.

[jeffreyH]

According to quantum mechanics neither space nor time are quantized. Only energy has quanta. From the continuums of space and time the magnitude of the energy quanta can be derived

[geordief]

Is the object not just following its trajectory in curved spacetime ?
What would its trajectory look  like  if it was modeled    in this way?

Would it not be similar to an object moving on the surface of the Earth with the difference that the latter is being constantly accelerated out from the centre of the Earth whereas the stone is in freefall immediately following its initial upward acceleration?

I wonder whether this "infinitesimal pause"  might be modeled as  simply a point on a geodesic in the curved spacetime model and the question not arise. 

[Colin2B]

I wonder whether this "infinitesimal pause"  might be modeled as  simply a point on a geodesic in the curved spacetime model and the question not arise. 

Firstly, it isn’t an infinitesimal pause it is genuinely a dt=0 at v=0. It is only infinitesimal if you choose to zoom out and sacrifice accuracy.
Yes, it can be modelled as a point on the geodesic, but that is exactly the same as modelling the point on the apex of the trajectory curve as described above, just a different perspective.

[Bill S]

Firstly, it isn’t an infinitesimal pause it is genuinely a dt=0 at v=0. It is only infinitesimal if you choose to zoom out and sacrifice accuracy.

I've not waded through this thread, so the answer may be in there somewhere.

 Are you saying that the stone does not actually stop?  Would that mean that the -ve > +ve acceleration happens at such low speed that the pause is only an illusion?

[Bill S]

I think the answer is : it spend at that point zero time

Just as an aside; what is the difference between being at a point for zero time, and not being there at all?

[Colin2B]

Are you saying that the stone does not actually stop?  Would that mean that the -ve > +ve acceleration happens at such low speed that the pause is only an illusion?

The answer is in @evan_au post #5.
Yes the stone does not stop because stopping implies spending time at a location in space. Yes, the pause is an illusion, as you say due to the low speed, also meaning the transition between up and down motion is seamless.
To suggest otherwise would mean there is a flat spot at the apex, and clearly there isn’t.

Just as an aside; what is the difference between being at a point for zero time, and not being there at all?

There is one for the philosophers.
However, it passes through that point so it must have been there 
Or does it just kiss the apex and move on 

EDIT: PS Here is another thought for you @Bill S - is it possible to say that any moving point (eg the tip of your nose as you walk down the street) spends any time at a particular point in space. If it does it is stationary. So, has your nose been anywhere?

[Bored chemist]

You can never divide by pure zero.

it's just my personal explanation.

Thanks for clarifying your lack of understanding.

[Bill S]

So, has your nose been anywhere?

Assuming that walking produces a continuous motion in a constant direction, it would seem that your nose does not stop.  Therefore, it spends no time at any specific point.  If every point along the way is designated as “a specific point”, it follows that your nose has spent no time at any point. So, if spending no time at a point (a) is the same as not being there (b), your nose has been nowhere.  Since we know that your nose has been somewhere, there must be a difference between a and b. 

So far, so good; but how do you define the difference?  Consider the following: If points A, B and C are contiguous points along your way, there is no space between them.  Your nose spends no time at A, B or C. Between A>C there is nowhere for your nose to be, so how long does it take your nose to move from A to C.    If you extend this “journey”, say, from A to Z; does that make any difference to the time taken?

[Bill S]

Another thought: If it takes your nose no time to travel from A to Z, it must travel at "c".  In order to do that, it must be massless.

There's no accounting for hooters. 

[geordief]

"Keep a clean nose. Watch the plain clothes. You don't need a weatherman. To know which way the wind blows."

The model  is not the same thing as what is observed.It is like Aesop's  tortoise trying to catch up with the hare

[Colin2B]

"Keep a clean nose. Watch the plain clothes. You don't need a weatherman. To know which way the wind blows."

Look out kid
It's somethin' you did
God knows when
But you're doing it again

Hare today, gone tomorrow

Another thought: If it takes your nose no time to travel from A to Z, it must travel at "c". 

Or instantaneous?