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When velocity=0 during maximum height when an object is thrown up, what does that actually mean?? does that actually mean that the object doesn't change it's position over dt time period when at maximum height?basically it hangs where it is for the dt time period?if so, then why doesn't acceleration cause it to have a infinitesmal change in position during this dt time interval?or, it does have a change in position but that is so small compared to dt time interval that we assume it to be 0,hence getting v=0 as well? Basically I know the maths behind this but i want to understand what actually happens in this scenario

Please keep in mind that I'm an elementary level student and please refrain from using high level language, I would be grateful if you can go as deep and fundamental explaining everything as possible while speaking in layman or simple terms

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  • $\begingroup$ It means that object is not moving at that instant. $\endgroup$
    – d_b
    Commented May 12, 2023 at 17:18
  • $\begingroup$ Why? Doesn't acceleration(gravitational acceleration g) act upon the object during that instant? Shouldn't it cause the object to have infinitesmal displacement??? $\endgroup$ Commented May 12, 2023 at 17:19
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    $\begingroup$ Yes, there is an acceleration. An object can be accelerating even if it is not moving. Note that we are not talking about a time interval (finite or infinitesimal), we are talking about a single instant in time. I would also say that I think you are confusing yourself with this idea of 'infinitesimal displacement' which I don't think is a useful concept here. $\endgroup$
    – d_b
    Commented May 12, 2023 at 17:26

5 Answers 5

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This is the kind of question that is better answered with an appeal to mathematical unity.

Draw the velocity time graph. You should get a straight line graph from positive velocity at initial time, until the velocity is negative. The line passes through a point in time where the velocity is zero. The line is continuous. It would be mathematically absurd for the velocity to be continuously transitioning from upward to downward without accepting that, at some time, it is zero.

The slope of the graph is the acceleration. It is downwards and constant. The little time $\mathrm dt$ does not cause an infinitesimal change in position. By definition of acceleration, that little time causes an infinitesimal change in velocity, so that it changes from zero velocity to negative velocity. Everything is fine.

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Good answers are provided, I only want to address your statement that:

i want to understand what actually happens in this scenario

To do that, you would have to conduct an experiment. While we model this situation with nice smooth functions such as, in this case, $v=v_0-gt$, in the real world, you would have to do something like the following:

  1. Record as accurately as you can, the heights $h_1$ and $h_2$ of the object at two nearby instants of time $t_1$ and $t_2$
  2. Calculate the average speed over this interval $$\bar{v} = \frac{h_1-h_2}{t_1-t_2} = \frac{\Delta{h}}{\Delta{t}}\tag{1}$$ to get a good approximation of the speed the object obtains during this small time interval.

(The bar over $v$ means average, not a vector)

Now, suppose you have conducted such an experiment, and you have a big notebook full of data entries that look like:

$$ (h_1, h_2) = (4.432m, 4.438m)\\ (t_1, t_2) = (1.21s, 1.235s)\\ (h_1, h_2) = (4.438m, 4.47m)\\ (t_1, t_2) = (1.235s, 1.295s)\\ (h_1, h_2) = \dots $$

You may ask yourself now the question, is there a simple way to find an entry for which the above calculation $(1)$ yields $\bar{v}=0$?

So first, we notice that $\bar{v}=0$ only when $h_1=h_2$. What does that mean? It means that if we want to record accurate readings of the heights and times corresponding to when the object reverses its direction, we need a height sensor that is sensitive enough to be able to resolve small differences in heights, obtained by the object at nearby instants of time, since clearly, as the object reaches the maximal height, it traverses smaller and smaller heights at equal (even if also small) intervals of time, which is just a fancy way of saying it slows down! :)

Now, if the height sensor's resolution is quite crude, and can for example only detect a position change in "chunks" of $10\text{cm}$, we may make a measurement in which the object's speed will appear to be zero $\bar{v}=0$. For example, if the object rises $3\text{cm}$ in tenth of a second ($0.1s$) we clearly won't be able to measure the speed correctly for that small time interval. Our readings of speed will only reflect averages taken over height intervals of $10\text{cm}$. So it's important, when plugging experimental data into an equation such as $(1)$, to take into account the resolution of our instruments. Getting $\bar{v}=0$ doesn't mean we actually managed to measure "the instant in which the speed vanishes" if we have a high resolution clock but a crude height sensor. The point here is that our equations, even when theoretically correct, can't give us good results if we feed them with poor data.

A similar thing happens if our clock resolution isn't very good: the denominator $\Delta{t}$ will become zero and the speed will be undefined. This simply reflects the fact that we can't compute an average speed over a time interval that we measured to be zero, and it just means that our instruments can't resolve this small time interval correctly.

So we see that we need to take care to only average over time intervals in which both our clock can resolve time correctly, and our height sensor can resolve distance correctly, so that $\Delta{h} \neq 0$ and $\Delta{t} \neq 0$. If we do that with reasonably good* instruments, plug our results into $(1)$, and plot the obtained $\bar{v}$ values as a function of the measured time $t$, we will get a graph similar to the following:

enter image description here

The above graph comes from a very primitive numerical simulation of such a situation, where we've thrown an object upwards. Each time interval is a thousandth of a second ($0.001s$) long, and the speed for each instant is plotted vertically. With real experimental data not all time intervals will be necessarily exactly equal, so this is still obviously an idealization. However, as you can see, we never actually measure the exact instant in which $\bar{v}=0$, but nevertheless such experimental data implies the object does momentarily obtain actual speed of $v=0$, despite the fact that we can only measure this directly to the degree that our instruments allow us. In fact, it can happen that in a certain experiment of this kind we will get $\bar{v}=0$ identically, not because of lack of resolution of the height sensor as I've mentioned before. I'll leave you to think about how that can happen exactly, but the point is, even then we are only measuring the average speed, never an "instantaneous" speed.

To summarize: Instantaneous speeds belong to the realm of math and calculus, while average speeds is what belongs to the realm of experimental Physics. But like in many other cases, experimental facts make us trust that "the one realm approaches the other" and we are doing the right thing when we model this and very many other situations via nice "smooth" functions such as, in this case, $v=v_0-gt$.

Full disclosure: I have near to zero experience in Experimental Physics, so I hope I am not making any gross oversimplifications here. I only wanted to convey some sense of what it would take to actually try to observe and demonstrate this phenomenon of speed vanishing experimentally.

*What are "good" instruments is ultimately decided according to the accuracy to which you want to perform your experiment

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$v=\frac{dx}{dt}:= \lim\limits_{\Delta t \to 0} \frac{x(t+\Delta t)-x(t)}{\Delta t}$

Reading the expression in words:

"Velocity equals the rate of change of position with respect to time, which equals by definition the limit, as time interval Delta t goes to zero, of the difference of the value of x at coordinate: 't plus Delta t' and the value of x at coordinate: 't', divided by Delta t."

Velocity is zero at those points for which, in the limit as you make the time interval smaller and smaller, you find that the change in the position over that time interval is either always zero or always decreases by a bigger fraction of its last value than the time interval decreased compared to its last value.

A graphical way of visualizing the previous paragraph is to imagine zooming in to an arbitrarily high zoom setting on a point on the curve. If the curve at that high zoom setting looks like a horizontal straight line, the rate of change is zero at that point.

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Velocity is displacement over time. If we take a point $P_a$ a little bit after the maximum height, it is always possible to find another one $P_b$ a little bit before it, so that both $P_a$ and $P_b$ have the same height. And no matter how close are $P_a$ from the maximum height, that procedure is always possible.

That is the meaning on saying that in the limit there is no displacement over time at the maximum height, or that the velocity is zero at that point.

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The energy E is conserved :

$$E=m\,g\,h+\frac 12 m\,v^2=\text{constant}\tag 1$$

you start with $~h=0~$ and the velocity $~v=v_0~$

and if the velocity $~v=0~$ then $~h=h_{v=0}~$

you obtain from equation (1)

$$ \underbrace{m\,g\,h_{v=0}}_{\text{potential energy }}=\underbrace{\frac 12 m\,v_0^2}_{\text{kinetic energy}}$$

this means that if the velocity $~v=0~$ the potential energy is equal to the starting kinetic energy


with

$$h=v_0\,t-\frac g2\,t^2$$

the acceleration is: $$\ddot h=-g\ne 0$$

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  • $\begingroup$ $-mv^2/2$ rather than $+$... a typo? $\endgroup$
    – Amit
    Commented May 16, 2023 at 15:18
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    $\begingroup$ @Amit you are right , it should be plus $\endgroup$
    – Eli
    Commented May 16, 2023 at 15:24

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