How can I find the angle of a projectile from its velocity and equation?

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The discussion focuses on calculating the angle of a projectile given its velocity and trajectory equations. The primary equation discussed is derived from the Pythagorean theorem, represented as \(10^2 - \left(\frac{15g}{v_z + \sqrt{v_z^2 - 8g}}\right)^2 = v_z\). Participants encounter challenges with imaginary numbers and algebraic manipulations while attempting to isolate \(v_z\) and \(v_x\). The consensus is that the problem may be physically impossible with the given parameters, particularly the initial velocity of 10 m/s.

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  • #31
Set the original equation equal to 0 i.e. f(Vz)=0 graph the equation on a graphing calculator or using some other graphing software. Determine the root where the plot crosses the x-axis. I have done so and the root is Vz=100.
 
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  • #32
I'll try this post again, doesn't look like it showed up.

Set the original equation equal to 0 i.e. f(Vz)=0. graph the equation on a graphing calculator or other suitable graphing software. Determine the root where the graph crosses the x-axis. I have done so and the solution is vZ=100 m/s
 
  • #33
That is weird, my calculator shows error, are you sure you did it right?
 
  • #34
After regraphing Vz=99.45 m/s not 100 m/s

I have graphed the equation on both a ti-83 calculator and advanced grapher software, both show roots of 99.45. You may want to change the order of the eqn to 100-X-(...) before graphing.
 
  • #35
That means that v_x is a negative?
 
  • #36
Hey, that result fits! Thanks!
 
  • #37
Can you tell me the values for the x and y-axis and the scale you used? My window doesn't have lines, so maybe I put the scale wrong.
 
  • #38
Hi IBY,

IBY said:
Hey, that result fits! Thanks!

I was just looking at this thread, and I guess I am confused. I thought the initial speed had to be 10 m/s? If so, then a z-component of 99.45 m/s for the initial velocity will not be a solution to this problem. Did I misread your early posts?
 
  • #39
It is 10m/s, but I had to find initial V for the z and x-axis to find the angle of 10m/s. Therefore, v_x^2+v_z^2=10^2 And v_x=\frac{15g}{v_z+\sqrt{v_z^2-8g^2}} Then, I have to use those values to find cotan (v_z/v_x) Which is the angle at which 10m/s goes. Yeah, you know what? I am having a bit of trouble.
 
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  • #40
IBY said:
Oh, and actually, it is \sqrt{99.45}, yeah, forgot to tell about that last part.

No, I don't think that works if I read the problem right. With that speed for vz, then vx would be about 0.74 m/s. So it would take about 20 seconds for the particle to reach x=15, and at that time it would be very far from y=10. Or am I misunderstanding the problem?
 
  • #41
I just figured out what I did wrong. It turns out that if I plug 99.45 in v_z, then the result of v_x is 0.75, which it turns out to be almost exactly 0.75^2+99.45=\sqrt{100}
 
  • #42
IBY said:
I just figured out what I did wrong. It turns out that if I plug 99.45 in v_z, then the result of v_x is 0.75, which it turns out to be almost exactly 0.75^2+99.45=\sqrt{100}

I'm not sure what you're saying here. (I'm sure the square root over the 100 is just a typo.) Are you agreeing that it is not a solution?
 
  • #43
Yes, I am agreeing. Oh, and the square root was the result of pythagorean theorem thingy. It is actually 10^2
 
  • #44
If I am understanding this problem, I don't think it has a solution. I think you can see that by examining the trajectories. For example, just based on the coordinates, what lower limit can you place on the angle?
 
  • #45
No limit, but if it has no solution, then I guess that puts an end to this problem, so all I need is like a confirmation.
 
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  • #46
Well, for example, you can determine that the angle must be at least 62.3 degrees, or else the particle will never make a vertical displacement of +4m (from y=6 to y=10). At that angle, the apex of the parabolic path is about 4.2m.

This means that if the particle goes through the point you want, it must be doing it on the way down.

Now think about what happens as the angles increases. What happens to the apex of the parabola? If you look at where the parabola crosses the y=10 line, the x-coordinates of those intersections are symetric about the x-coordinate of the apex. Does that make sense?
 
  • #47
Ok, yes, except the part where you said the particle must go down. But, is the problem even solveable? If that was what you were trying to say, then I must not have understood it at all.
 
  • #48
The apex of the parabola is the particle's highest point, right? So at the angle of 62 degrees or so, the parabola just barely touches the y=10 line we want. And it does it at x=4.2.

As the angle increases, the apex gets higher, but also closer to the y-axis. So the x-coordinate of the apex get smaller, which means the apex is moving away from the (x=15,y=10) point we want the particle to reach. So if it made it to that point, it must be doing it after the particle has already passed the highest point.

Now here's the idea: if the largest x value of the apex we're allowing is 4.2, then past x=8.4, the particle must definitely be below the original starting height. So past x=8.4, the y-displacement must be negative, so there's no way it can reach x=15 and be above its starting point.
 
  • #49
Oh, now I get it, thanks! Now I know that I have wasted over one week pondering upon it, when there was no answer. ^_^
 
  • #50
Well, I'm glad to have helped; and I definitely know how aggravating these things can be!

But trust me, when you think really hard about a problem for a long time, no matter how aggravating, the effort is rarely a waste.
 

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