# My possible misunderstanding on a freely falling object

Tags:
1. Sep 5, 2015

### Anama Skout

Consider the famous kinematics equations $$x_f=x_i+v_i\cdot t+\frac12at^2$$ My question is: does this equation work for every point A & B along the parabola formed when dropping an object?

2. Sep 5, 2015

### Staff: Mentor

Yes. Each point is represented by a different t. To make that clear, I would write it as:

$$x(t)=x_0+v_0 t+\frac{1}{2}a t^2$$

3. Sep 5, 2015

### Anama Skout

Thanks for your answer. But I have something different in mind. Suppose that the point A has $x_{A,0}$ and $v_{A,i}$, then the equation will have the form $$x_A(t)=x_{A,i}+v_{A,i}\cdot t+\frac12at^2$$ and for a point B which has $x_{B,0}$ and $v_{B,i}$ then the equation will have the form $$x_B(t)=x_{B,i}+v_{B,i}\cdot t+\frac12at^2$$ Will this yield the same values in that for instance if we use them to determine what will happen after $t=5$ as long as we know what $x_{A,0}$, $v_{A,i}$, $x_{B,0}$, and $v_{B,i}$ are (and the time difference between the two)?

4. Sep 5, 2015

### Staff: Mentor

Why don't you try it and see for yourself?

5. Sep 5, 2015

### Anama Skout

I tried it with three points, always worked well (except the times when I made some mistakes -- of course), however I'm more interested into a rigorous proof.

6. Sep 5, 2015

### Staff: Mentor

Should be doable with a bit of algebra. Go for it!

7. Sep 5, 2015

### Anama Skout

I know it must be a simple thing I'm missing, but I can't figure it out, here's my try. Consider two points A & B such that $x_{A,i}=R+x_{B,i}$ and $v_{A,i}\neq v_{B,i}$ (for generality), then I should show that
1. it implies that $x_{A,f}=R+x_{B,f}$
2. it implies that the final velocities wont change (how do I formulate that?)
I arrived at the following equations $$v_{A,f}-v_{B,f}=v_{A,i}-v_{B,i}\\x_{A,f}-x_{B,f}-R=t(v_{A,i}-v_{B,i})$$
And for that equality to happen we must have $v_{A,i}-v_{B,i}=0$ but this contradicts our assumption that $v_{A,i}\neq v_{B,i}$, since clearly if -- for example -- you have an object falling onto the ground, if you take two distinct points in its trajectory, the velocity at each point wont be the same.

So what am I missing?

(here's a visualization of the situation)

File size:
6.4 KB
Views:
39
8. Sep 5, 2015

### Staff: Mentor

The given form of the equation only works for $x_0$ and $v_0$ defined at $t=0$. At $t=0$ the object is only at one point. If you use the same formula for a point which is on the parabola at a different time then you are going to get a different parabola which has that point at $t=0$.

In other words, unless $A=B$ the equations that you have written represent two distinct parabolas. I.e. the first equation passes through $A$ at $t=0$ and the second equation passes through $B$ at $t=0$, they are not the same parabola even if both parabolas pass through $A$ and $B$.

Also, I am not sure if you are thinking about parabolas through space or parabolas in spacetime. The equation you have written is for 1D motion, so it is only a parabola in spacetime.

9. Sep 5, 2015

### Anama Skout

Ah, forgot about time, I should have added the following constraints $t_B=T+t_A$, but again this doesn't yield much...

10. Sep 5, 2015

### Staff: Mentor

So are you thinking of parabolas in space parametrized by time or are you thinking of parabolas in spacetime?

11. Sep 5, 2015

### Anama Skout

I'm thinking of parabolas in space parametrized w.r.t time. (like the one I drawed)

Last edited: Sep 5, 2015
12. Sep 6, 2015

### Staff: Mentor

Why don't you treat each axis separately? I would think that would be simpler.

13. Sep 6, 2015

### Anama Skout

I only treated the system w.r.t to a coordinate system in which the $x$-axis is vertical.

14. Sep 6, 2015

### Staff: Mentor

Ah, so you are only talking about the vertical axis (as I first thought) despite the mention of parabolas. OK.

15. Sep 6, 2015

### Staff: Mentor

I am not quite sure why you want to put yourself through this, but OK. So a parabola in space can be defined by the parametric equations:
$x(\tau)=p \tau^2 + h$
$y(\tau)=2 p \tau + k$
$z(\tau)=0$
Where the vertex of the parabola is at $(x,y,z)=(h,k,0)$, and $p$ indicates how flat the parabola is. The x direction is vertical, the y direction is horizontal along the arc of the parabola and the z direction is perpendicular to the parabola.

$\tau$ is not time, it is a somewhat arbitrary parameter. To use this approach then you would need to solve for the relationship between $\tau$ and $t$. That would involve some more free parameters which would allow you to have the same spatial parabola represented by the same equations in $\tau$ independently of when the object passes through any given points on the parabola.

I leave the details of figuring out the relationship between time and $\tau$ to you.

16. Sep 6, 2015

### Hamza Abbasi

How would this equation work for parabola? This is a straight line equation isn't it?

17. Sep 6, 2015

### Staff: Mentor

The motion is along a straight line. But expressed as a function of time, the equation describes a parabola.

18. Sep 6, 2015

### Chandra Prayaga

The confusion here is that you have defined too many variables. Point i or 0, is a chosen initial point on the trajectory. A and B are two other points on the same trajectory. So there is no separate initial position for A and a separate one for B.

19. Sep 7, 2015

### Hamza Abbasi

E
Equation is of straight line but it explains parabola?? How come?

20. Sep 7, 2015

### Staff: Mentor

If you plot x as a function of time, it is not a straight line.