# A question about gravity/upward motion

1. Dec 20, 2008

### Zoideberg

Hi.

Me and my friends have been debating a question for years.

If you throw an object into the air does it stop for any period of time?

My reasoning has been that it doesn't. Of course it stops when it stops moving upward, before moving downward, but it does so for 0 seconds. At the exact moment the upward force is gone and the object "stops", gravity starts pulling the object down. If it would stop for any period of time in mid air (i.e. longer than 0 sec.) it would be in zero gravity.

Can anyone clear this up?

Thanks,
R.

2. Dec 20, 2008

### Staff: Mentor

Clear what up? Sounds like you understand things well enough. The vertical component of the velocity is constantly changing.

3. Dec 20, 2008

### Zoideberg

I guess the question should rather be, can someone explain this in a more "formal" way in connection with the laws of gravity.

4. Dec 20, 2008

### Staff: Mentor

I don't understand what you're looking for. Do you agree that the object is under constant acceleration due to gravity? (Ignore air resistance.) So the velocity is constantly changing.

Say you toss it straight up at 5 m/s. Does it ever have a velocity of 0 m/s... or 4 m/s? Sure--for an instant. For any "length of time"? No.

Is this an argument about semantics? As in: If it isn't at rest for any length of time, then it is never at rest! If that's what this is about, then explain that in common usage velocity is the limit of Δx/Δt at any given point.

5. Dec 20, 2008

### Nick89

This is a misunderstanding, although not completely related to your question. The upward force is gone as soon as you release the object. Gravity is pulling the object down all the time, not just when it is moving downwards.

When you are holding the object stationary, gravity pulls it down, but your arms exert a force of equal amplitude upwards, so the object remains stationary. When you throw it, for a short amount of time your arms exert a force greater than the force of gravity, that is why it accelerates upwards. As soon as you release it however that force is gone, and the only force acting on the object is gravity (ignoring air resistance). That is also the reason is slows down and falls back to earth: the object already had an upward velocity, but gravity slows it down until it stops, then it falls back.

If you are not throwing the object exactly upwards, then the path the object follows is a parabola. A parabola has one point where the slope is zero (at the 'top') and that is exactly where the object is stationary. The top of the parabola is one point, it does not exist at multiple moments in time, therefore the object does not stay stationary for any amount of time.

6. Dec 20, 2008

### Staff: Mentor

Good catch, Nick. (I missed the implications of that statement on first reading.)

7. Dec 20, 2008

### HallsofIvy

The acceleration is always non-zero (in fact, it is -9.81 m/s2) so speed is always changing. It cannot be 0 (or any constant value) over any interval. I think that is what you are looking for.

8. Dec 20, 2008

### Gnosis

You are correct. The object has no “hang time” at 0 m/s, as such a feat would require gravity to be overcome. The instant the object’s upward motion has decreased to 0 m/s, the object instantly begins its free-fall descent toward the Earth’s center of mass no different than if you had released the object by hand from a 0 m/s velocity.

9. Dec 21, 2008

### DaveC426913

Note that "the point at the top" of the parabola of motion is arbitrary and entirely dependent on the observer's frame of reference (we just conventionally use the Earth's surface as the FoR). You can change the observer's frame of reference (say, by having him moving upward at some consant rate) and this object's apex will be observed a different time and place.

It follows then that the apex of the parabola is no more significant than any other point on it. An object no more stops there than it does anywhere else on its path.

10. Dec 21, 2008

### schroder

I think this question boils down to a definition of instantaneous time and a period of time. There really is not much physics involved. The object passes through an instantaneous velocity of zero, but does not have zero velocity throughout any interval of time.

I suppose that someone could argue that an “instant” however small still represents an interval, just a very small interval. But that would be incorrect, as far as the mathematical concept of an instant is concerned. But you should not be surprised that you and your frinds have had this disagreement for “years”; some of the greatest mathematical minds argued over this for over a century until Augustin-Louis Cauchy finally set everyone on the right track with his definition of a Limit. Still, while the concept of a limit itself is quite simple, the definition is much harder!

11. Dec 21, 2008

Geee .. you can claim according to some unwritten book of blah blah that an instant of time doesn't exist . But it seems to me that this is simply false in math .

look at the curve of marginal speed .. speed could be seen as the slope of a tangent to the curve of a parabola .

At the Zenith , the tangent to the curve is flat . Thus the speed i zero .

If you want to discuss the quanta of time then yes oK lets do that .. but to claim like that that there is no instant of time .. tsk tsk .. someones been taking themselves far to seriously :)

12. Dec 21, 2008

### Hootenanny

Staff Emeritus
The OP makes a valid point and isn't "taking themselves fair to seriously". If you would like to put a more mathematical slant on it, I can do that. The speed of a ball that has been thrown vertically upwards may be written

$$v\left(t\right) = v_0 - gt$$

Of course v(t) may be instantaneously zero since it has the root t = v0/g. However, what the OP wants to know is:

Does there exists a non-empty set S={t : t>0 | a < t < b} such that

$$v\left(t\right) = 0 \; \forall \; t \in S$$

Is that still "false in the math", whatever that means?

Last edited: Dec 21, 2008
13. Dec 21, 2008

### valleyman

I completely agree.
This brings up to my mind the old Zenon paradox, even if here we are talking about acceleration and velocity, instead of velocity and space

14. Dec 21, 2008

### mordechai9

We have the position, velocity, and acceleration of the ball. I'm sure you have a basic understanding of these notions. Well, when we say that an object "stops", we mean that the velocity is zero and the acceleration is zero. Otherwise, the object would be moving, or it would be in the process of moving (accelerating).

Now let's consider your example. While the velocity is zero at the apex, the acceleration is nonzero, because the force of gravity is accelerating the object, even at that instant. Hence, by the previous paragraph, the ball does not stop.

15. Dec 21, 2008

### Gnosis

There can be no argument that the upward velocity of an object in free-fall will finally decrease to precisely 0 m/s, as it must prior to its direction change however, there is no duration for which the object remains at 0 m/s. There is only the instant in recordable time where the object achieved 0 m/s via a stopwatch or freezable real-time clock, but the object itself continues its accelerated motion and descends.

Example:

If we launched a ball vertically from the Earth’s “surface” in an air evacuated tube at a velocity of precisely 19.6 m/s, a precision timing device would indicate that the ball achieved 0 m/s in precisely 2 seconds via the equation below:

19.6 m/s / 9.8 m/s^2 = 2 seconds

The ball would also require precisely 2 seconds to free-fall back to the Earth’s surface.

If the ball were able to “hang” (retain some time duration at 0 m/s), then it would require more time than 2 seconds to free-fall back to the Earth’s surface therefore, greater than 4 seconds total flight time up and back down. Countless tests reveal there is no length of time that the object (the ball in this case) remains at 0 m/s.

Additionally, acceleration graphs demonstrate that a velocity of 0 m/s does in fact occur, but that moment is per a precise moment, not per a time duration.

Another example:

You release a ball from a motionless arm to free-fall in Earth’s gravity. Ask yourself, “Does the ball linger at 0 m/s for ANY interval of time once it has been released?” The answer of course, is no. The instant the ball is released to free-fall, its 9.8 m/s^2 accelerated descent commences. Even if measured in the millionths or billionths of a second after its release, the ball has begun to accelerate and descend. Of course, at these miniscule time intervals, the ball won’t have fallen but a miniscule distance, nor will it have achieved any significant velocity, but it WILL have begun its descent and it WILL have gained velocity no matter how infinitesimally small.

16. Dec 21, 2008

### DaveC426913

Not wanting a harp on the point but my argument in post#9 demonstrates that the point at the top of the parabola is no more special than any other point on the arc.

It's intuitive that a ball does not, even for a short duration, have a constant speed anywhere on its upward or downward path. It is then no leap to see that the apex is no different.

17. Dec 21, 2008

### Integral

Staff Emeritus
The apex is special in that at that instant the velocity and therefore kinetic energy are zero. This makes a great place to catch things.

As Dave said, it spends as much time at zero as it does any other velocity between 0 and the initial velocity.

18. Dec 21, 2008

### DaveC426913

True, but only wrt to the ground which, while a very convenient FoR, is physically and mathematically arbitrary.

19. Dec 21, 2008

### Gnosis

Respectfully, the original post wasn't questioning the velocity while traveling upward or downward. He specifically asked, “If you throw an object into the air, does it stop for any period of time?” His reasoning was that it does not, and he was quite correct.

Additionally, the apex is quite unique. It is the point where an object’s vertical velocity reaches precisely 0 m/s, it is the point where vertical direction changes, and it is the point where vertical kinetic energy has diminished to zero.

Lastly, if every point of the curve were as intuitive as you seem to think, then no one would have cause to present the question in the first place. Clearly, it is not as intuitive to all, as you have suggested and that is why the question has been presented.

20. Dec 22, 2008

### Nick89

I think what Dave is trying to say is that the (position of the) apex will change depending on which frame of reference you choose. That is obviously true, but if you choose another frame of reference, the velocity of the ball will be zero somewhere else, again at the apex of the (new) curve.