My possible misunderstanding on a freely falling object

[nasu]

No, the difference is 20.4 m.
A has 0 coordinate and B has 20.4 m. You are mixing coordinate and time.

Because the speed is the derivative of the position in respect to time. Adding a constant to the coordinate (this is what a translation means) does not change the derivative.

 

[Anama Skout]

No, the difference is 20.4 m.
A has 0 coordinate and B has 20.4 m. You are mixing coordinate and time.

Because the speed is the derivative of the position in respect to time. Adding a constant to the coordinate (this is what a translation means) does not change the derivative.

Oh, sorry again for the mistake. Thanks for that, now it become all clear!

 

[Anama Skout]

But just one last thing, how can I show that the speed will be the same without using calculus?

 

[nasu]

Well, look at finite changes. You can think about an average speed, which is Δx/Δt.
Calculate the change in position for two different origin.
For example, if the stone was at D and it moves 1m below D, by how much increases the distance between the stone and C or B?

 

[Anama Skout]

Well, look at finite changes. You can think about an average speed, which is Δx/Δt.
Calculate the change in position for two different origin.
For example, if the stone was at D and it moves 1m below D, by how much increases the distance between the stone and C or B?

I get this $$\overline{v}_1=\frac{\Delta x}{\Delta t}=\frac{(x_B-x_D)-(x_B-(x_D-1))}{t_{B;D}-t_{B;D'}}=-\frac1{t_{B;D}-t_{B;D'}},$$ where $t_{B,D}$ means the time the particle is at D relative to B, and D' stands for the point when the stone moves 1m below D. And with respect to C $$\overline{v}_2=\frac{\delta x}{\delta t}=\frac{(x_C-x_D)-(x_C-(x_D-1))}{t_{C;D}-t_{C;D'}}=-\frac1{t_{C;D}-t_{C;D'}},$$ so to show that these two things are equal one would require that $$t_{B;D}-t_{B;D'}=t_{C;D}-t_{C;D'}.\tag1$$ I'm stuck here...

 

[nasu]

These indexes on time don't make much sense to me.
What would you mean by t_C;D for example?
It cannot be the time to move from C to D because this is irrelevant here. It's not about the stone thrown from the building.
Just imagine that you have a stone standing at D and then you move it down by 1m.

The time does not depend on what origin you choose for coordinates.
If the stone moves from D to D' in Δt, then this is the time interval for that motion.
If you measure distances from B or from C, the time interval is still Δt and the stone still moves from D to D'.
Your clock does not "care" how you measure the distance D to D'.

Only in relativistic regime the time may depend on the coordinate systems, if these are moving relative to each other. But here all these points (A,B,C,D) are not moving.

 

[Anama Skout]

What would you mean by tC;D for example?

As I explained before $t_{B,D}$ is the time the rock is at D with respect to B.

 

[nasu]

The rock is not at D for some finite time. It may pass through D or start from D. In any cases, it is not at D for some time. Unless is at rest there for some time. But this time won't be relevant for any of the questions discussed here.

You mean time to travel from D to B? Then this is irrelevant for the question regarding the invariance of velocity. The situation I described is a rock which is at D and then moves 1 m away. That rock was never at B so there no time to travel from B to D.

So I don't know, you seem to have some confusion about time but I cannot figure out what exactly is.
Think about it, 1 minute (let's say between 1:00 PM and 1:01 PM is the same no matter if you start driving from New York City or from London or any other city. Or anywhere on Earth. This is why we can use time shown by clock to correlate our activities.
When you say it took you 30 minutes to do something people don't ask from where were you going or something like this.

 

[Anama Skout]

I mean, look at the previous example, the time with respect to C was $t=(5−4.08)\;\rm sec=0.92\;sec$ and with respect to B it was $t=(5−2.04)\;\rm sec=2.96\;sec$..

 

[nasu]

The "time with respect to C" or B is meaningless. C and B are points in space and not events. The origin of the time should be some instant in time, some event. Like you turn on your stopwatch or the rock was thrown up when the clock showed 12:00 PM.

The only way to make some sense of this would be to use the original rock as a clock and take different origins of time as the instants when the rock passes these points.
For example, we look at an object (not the rock) moving from D to D' in two different ways:
1. Use D as origin of coordinates and the time the rock passed through D as origin of time.
So the change in position will be Δx=1m=x_D-x_D'
The time it takes to move from D to D' is Δt, which is t_D' -t_D
where t_D' is the time when the object was at D' and t_D the time it was at D, as measured from the instant the original rock passed D. Which could have happened yesterday or last year, it does not matter. Only the difference matter.

2. Use C as origin of coordinates. Use as origin of time the instant the rock passed C.
We know that C is 22.5 m higher than D. So the coordinates of D and D' will each have an extra 22.5 added, when compared with the same coordinates in part 1.
So when you take the difference x_D-x_D' the 22.5 will cancel out and you get the same Δx.
The time the rock passed C is 0.92 s before it passed D. So when you calculate Δt as t_D' -t_D for this new initial moment, each term will have an extra 0.92 s and when you take the difference, it cancels out too. So both Δx and Δt are the same as before and the velocity will be the same.