# Projectile Motion with multiple variables

1. Nov 21, 2013

### frensel

1. The problem statement, all variables and given/known data
As shown in the figure below, a particle is moving in a circle of radius $R$ with constant speed $v$. At some location, the particle is detached from the circle and falls with a parabola path to point A. What is the horizontal range $x$ of the projectile?

2. Relevant equations
Writing the kinematic formula in component form, we have
$$x=v\sin(a)t$$
$$h-y=v\cos(a)t+\frac{1}{2}gt^2$$
and using Pythagorean theorem, we get
$$R^2=x^2+y^2$$
Since $\sin(a)=y/R, \cos(a)=x/R$, the first and second equations become
$$x=\frac{vyt}{R}$$
$$h-y=\frac{vxt}{R}+\frac{1}{2}gt^2$$
3. The attempt at a solution
I have tried to solve the above three equations, but they are too difficult to solve. Some cubic functions will appear and it seems too complicated to be a validated way to solve this problem.

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Last edited: Nov 21, 2013
2. Nov 21, 2013

### voko

Not correct, because it does not take into account the initial x position.

Not correct either. x and y may have such (well, similar) relations with a only when the particle is in circular motion. But once it has detached, these no longer hold true.

3. Nov 21, 2013

### frensel

In order to arrive at point A, the particle should be detached from the circular track at some point with fixed $x$ and $y$ and they are related to some angle $a$. While $x$ and $y$ are determined, we therefore can obtain the initial velocity of the particle and set the particle in parabola motion to point A.

4. Nov 21, 2013

### voko

This is not what your diagram says. $x$ is not fixed and it is not where the particle detaches, so $x = R \cos a$ is clearly wrong. $y$ is fixed, and that means the second equation you gave is incorrect. Your diagram and your equations are not consistent, which is probably because you confuse initial values with dynamical variables.

5. Nov 21, 2013

### frensel

Well...I can't understand your comment. $x$ and $h-y$ are the horizontal and vertical displacement from the initial position to point A. We want to know the final displacement of the particle, not the intermediate position. If $x$ and $y$ are determined, we get the initial velocity in horizontal ($v\sin(a)$) and vertical ($v\cos(a)$) directions.

Last edited: Nov 21, 2013
6. Nov 21, 2013

### voko

If $x$ and $y$ are not the initial position, then I will say again: this

is not correct.

7. Nov 21, 2013

### frensel

I think it is correct. Noticed that the direction of the initial velocity is perpendicular to the radius of the circle.

8. Nov 21, 2013

### voko

You said $x$ and $y$ were not initial positions, now you imply they are. The equations that I said were incorrect are not correct for all the values of $x$and $y$; they are correct only at the initial position; but you are looking for the final position.

9. Nov 21, 2013

### frensel

$x,y,a$ are not time-dependent. They are used to determined the initial velocity, therefore the projectile. For certain $x$, we can use $R^2=x^2+y^2$ and trigonometry to obtain $y$ and $a$. There is only one solution of $x$ (or $y$, $a$, and therefore initial velocity) so that the particle can arrive at point A.

10. Nov 21, 2013

### voko

In the very first equation in the very first post, $x$ is time dependent. So is $y$ in the next equation. You have been on and off saying they are or they are not time dependent. Fact is, your equations ARE CONTRADICTORY. Make up your mind, use proper notation, which does not confuse initial value with dynamical variables, and move on. You have made a lot more difficult than it needs to be, you do not need to waste any more of your time on this.

11. Nov 21, 2013

### frensel

Sorry, I still can't figure out how to write down the proper equations...
For the equation $x=v\sin(a)t$, it does not mean that $x$ is time-dependent. $x$ is the final horizontal displacement, therefore $t$ is the the final time. With some final time $t$, the final horizontal and vertical displacement are $x$ and $h-y$.
For instance, if I throw something with initial horizontal speed $v_0$ that its horizontal displacement are 5m, I can write $5=v_0t$. That doesn't mean 5m is time-dependent.

12. Nov 21, 2013

### voko

Look. In one message you say $x$ is the initial displacement. In another you say it is final. It CANNON be the same symbol in both cases, UNLESS we think it is time dependent. EVEN IF we think it is time dependent, then you still have the problem I mentioned in #6, some of your equations apply ONLY at the initial position. You can fix that by using other symbols for the initial position, say $X$ and $Y$, and then you will immediately see that your method of #1 does not work at all.

13. Nov 21, 2013

### frensel

I haven't said $x$ and $y$ are initial displacement... I said they are "horizontal and vertical displacement from the initial position to point A".

14. Nov 21, 2013

### voko

Yes you did. Quoting #3:
And you did not just say that, use wrote the corresponding equations in #1 and used them, ending up with a meaningless result.

I have repeated all this a few times now. If you want to prove that what you do is correct, then why are you even here? I am sure you know SOMETHING is not right with your method. I am telling you what is wrong. It is about time that you listened.

15. Nov 21, 2013

### frensel

I still didn't get it. It seems that it will not make any difference if I write down the equations in dynamical form.
I assume $x,y$ are dynamical variables, $X$ and $Y$ are final horizontal and vertical displacement at final time $t=T$. We have
$$x=\frac{vYt}{R}$$
$$y=\frac{vXt}{R}+\frac{1}{2}gt^2$$
$$R^2=X^2+Y^2$$
When $t=T$, $x=X,y=h-Y$, therefore
$$X=\frac{vYT}{R}$$
$$h-Y=\frac{vXT}{R}+\frac{1}{2}gT^2$$
They are the same old equations...

16. Nov 21, 2013

### voko

This is not what I suggested. I suggested that $X$ and $Y$ be the INITIAL positions of $x$ and $y$. You have a problem with the INITIAL positions, because you do not have any symbols to denote them.

17. Nov 21, 2013

### frensel

The initial position is $x=0,y=0$ at $t=0$, the final position is $x=X,y=h-Y$ at $t=T$.

18. Nov 21, 2013