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

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The discussion revolves around finding the angle of a projectile given its velocity and motion equations. The user struggles with the calculations involving vertical (Vz) and horizontal (Vx) components of velocity, leading to inconsistencies and imaginary numbers in results. Various approaches are suggested, including using the Pythagorean theorem and simplifying equations, but the user remains stuck on algebraic manipulations. There is a consensus that the problem may not be physically feasible with the given initial conditions. Ultimately, the conversation highlights the complexities of solving projectile motion problems analytically.
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Find angle of projectile from equation 15*9.8
(-------) + Vz^2 = 10^2
^ Vz+√Vz^2-8*9.8
Vz^ / It is a 2d motion problem.
| / Find Vz, then Vx (which is the left term, the fraction one)
| / 10m/s to find angle at which projectile was shot.
| / To do check, I basically did the above equation into
| / Vz=10^2-Vx^2, but no matter what I do, my results
|)__________>Vx of Vz always ends up in the end making Vz^2+Vx^2
not equal 10, and always greater than 10 in my three days attempt at this, and in the end, it turns into a jumble of mess, as I have also turned up with imaginary numbers in the middle of the steps. So, the problem is, how to solve the above equation so that I can find angle by tangenting Vz/Vx, and is it supposed to come out with imaginary number? I am only slightly familiar with complex numbers, so I can't progress further,
 
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Welcome to PF IBY,
IBY said:
Find angle of projectile from equation 15*9.8
(-------) + Vz^2 = 10^2
^ Vz+√Vz^2-8*9.8
Vz^ / It is a 2d motion problem.
| / Find Vz, then Vx (which is the left term, the fraction one)
| / 10m/s to find angle at which projectile was shot.
| / To do check, I basically did the above equation into
| / Vz=10^2-Vx^2, but no matter what I do, my results
|)__________>Vx of Vz always ends up in the end making Vz^2+Vx^2
not equal 10, and always greater than 10 in my three days attempt at this, and in the end, it turns into a jumble of mess, as I have also turned up with imaginary numbers in the middle of the steps. So, the problem is, how to solve the above equation so that I can find angle by tangenting Vz/Vx, and is it supposed to come out with imaginary number? I am only slightly familiar with complex numbers, so I can't progress further,
I can't make head nor tail of your post. Could you please repost your question so that it can be read clearly. If you need to include a diagram you can attach an image, but please don't use ACSI diagrams as they rarely display correctly. If you need to use formulae you may correctly typeset equations using https://www.physicsforums.com/showthread.php?t=8997".
 
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Okay, this thing is not working as I want it to be, I will have to rework it.
 
Unclear.
 
Find angle of projectile from equation (V is velocity, x and z are the directions of the two velocities, x is to the right, and z is up)
10^2-(\frac{15*9.8}{v_z+\sqrt{v^2_z-8*9.8}})^2=v_z (the equation is basically a reworked pythagorean to find Vz)
It is a 2d motion problem. Find Vz, then Vx (which is the middle term, the fraction one)
to find angle at which projectile was shot at. To check, I basically did the above equation into V^2_z=10^2-V^2_x, but no matter what I do, my results of Vz always ends up in the end making V^2_z+V^2_x not equal 10, and always greater than 10 in my three days attempt at this, and in the end, it turns into a jumble of mess, as I have also turned up with imaginary numbers in the middle of the steps. So, the problem is, how to solve the above equation so that I can find angle by tangenting Vz/Vx, and is it supposed to come out with imaginary number? I am only slightly familiar with complex numbers, so I can't progress further.
 
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From what i could gather, you basically have to solve for Vz from the baove equation.right?

I think the xpression is pretty simple and can be easily solved by transposing Vz from rhs to lhs and making rhs equal to zero.Maybe first you put 'g' instead of 9.8 to make the problem much more pleasurable.Use simple maths identity a^2-b^2 and simplify.

Also going into the physics part, try to resemble the above expression to any of the forumulae given in your textbook related to projectile motion and your answer shall be much more simple.
 
Ok, but it doesn't work. Look:
10^2-(\frac{15g}{v_z+\sqrt{v^2_z-8g}})^2-v_z^2=0
I squared term x and 10.
100-\frac{225g^2}{v_z^2+v_z^2-8g}-v_z^2=0
Added like terms on the bottom of fraction and moved 100 to other side.
-\frac{225g^2}{2v_z^2-8g}-v_z^2=-100
Multiply the bottom of fraction to both sides.
(2v_z^2-8g)\frac{-225g^2}{2v_z^2-8g}-v_z^2=-100(2v_z^2-8g)

-225g^2-v_z^2=-200v_z^2+800g
Transferred two terms and converted g to 9.8 and solved rhs.
199v_z^2=800g+225g^2

199v_z^2=800*9.8+225*9.8^2

199v_z^2=29449
Divided 199 and sqrt to finally get
\sqrt{v_z^2}=\sqrt{147.98}
v_z=12.6

Now, I plug it in the original equation and solve.
(\frac{15g}{v_z+\sqrt{v_z^2-8g}})^2+v_z^2

(\frac{15*9.8}{12.16+\sqrt{12.16^2-8g}})^2+12.16^2

(\frac{147}{12.16+\sqrt{12.16^2-78.4}})^2+12.16^2

(\frac{147}{12.16+8.33})^2+147.87

(\frac{147}{20.49})^2+147.87

51.47+147.87

\sqrt{199.34}
Which is equal to 14.12, which is not equal 10, so, yeah, I am stuck, still.
 
You must also multiply the bottom term by -V2z.(fourth step)
 
How come?
 
  • #10
Algebraic principle...

ex. 100 - 5/5 = 99

100(5) - 5/5(5) = 99(5) correct


100 - 5/5(5) = 99(5) Incorrect
 
  • #11
Ach!*smacked my head* Of course. :)
 
  • #12
It looks like you made an algebra error in your first step. Look at the denominator of the second term of the l.h.s. in your first equation. When the fraction is squared, we have

<br /> (\frac{15g}{v_z+\sqrt{v^2_z-8g}})^2=\frac{225g^2}{v_z^2+2v_z\sqrt{v^2_z-8g}+v_z^2-8g}<br />

But either way, it looks like it'll still be really nasty to solve for vz. Can you explain how you initially developed your equation, ie how you used pythagorean thm?
 
  • #13
The problem was a projectile shooting from 6ft high, and the projectile hits a spot in coordinate (15,0,10)ft, and I had to find the velocity, then find the angle at velocity 10. Separating them, one gets
10=-\frac{1}{2}gt+v_zt^2+6 and 15=v_xt
In the second equation, v_x=\frac{15}{t}
In the first one, I moved 10 to the other side, making it
t=\frac{v_z+\sqrt{v_z^2-8g}}{g} (plus in the quadratic cause I want to choose the side when the ball hits the spot) I insert this in v_x=\frac{15}{t}, making initial velocity

(\frac{15g}{v_z+\sqrt{v^2_z-8g}},0,v_z)

Then the problem asked me to find the angle when it shoots at initial velocity 10. Therefore, I have to find v_x and v_z, so I did v_x^2 +v_z^2=10^2, and to find v_x, I have to find v_z. That is how I developed it.

By the way, how did you get that for the bottom of the fraction when it is squared?
 
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  • #14
Is distance in ft. or meters?

Also, there must be a time interval or a given angle.
 
  • #15
Meters, sorry. woops
 
  • #16
(a + b)^2 = a^2 + 2ab + b^2. In this case, a is v_z, and b is the square root
 
  • #17
To find the initial velocity we must be given a time interval or launch angle.
 
  • #18
The initial velocity is given, which is 10m/s. Angle is what I have to find by doing the cotangent with v_z and v_x, and to do it, I have to find both values.
 
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  • #19
Ok, knowing that makes this problem much easier.

If the coordinates are 15,0,10, than the total distance in the x direction is 18.03m
 
  • #20
No, that is not it. The coordinates you put out is for distance. What I need is velocity, which is v. I found the equations for velocity, then I have to use the equations I found to find v_z and v_x when those two combined is equal 10. Then after I find v_z and v_x, can I find the angle at which the initial velocity went at.
 
  • #21
I tried approaching the problem a slightly different way, and seeing as the details are mostly an exercise in math, I don't see any harm in laying some of them out. I started as you had,

<br /> (1)\:\:\:\:\:4=-\frac{1}{2}gt^2+v_zt<br />

We also have, in the x direction, that
<br /> (2)\:\:\:\:\:t=\frac{15}{v_x}

Substituting eqn (2) into eqn (1), we get
<br /> (3)\:\:\:\:\:4=-\frac{1}{2}g\frac{15^2}{v_x^2}+v_z\frac{15}{v_x}

In the problem, they're asking for the angle between vz and vx, or equivalently, the slope between vz and vx. Call this slope m, so that
<br /> m=\frac{v_z}{v_x}=tan\theta

If we substitute for m in eqn (3), we get
<br /> (4)\:\:\:\:\:4=-\frac{1}{2}g\frac{15^2}{v_x^2}+15m
or
<br /> (5)\:\:\:\:\:m=\frac{4}{15}+\frac{15g}{2v_x^2}

Okay now we look at the other criterion that the initial velocity be 10. That is,
v_x^2+v_z^2=10^2

Dividing by vx2...
1+\frac{v_z^2}{v_x^2}=\frac{10^2}{v_x^2}
1+(\frac{v_z}{v_x})^2=\frac{10^2}{v_x^2}
1+m^2=\frac{10^2}{v_x^2}

If we now solve for vx2 and substitute into eqn (5), we can solve for m, and thus find the solution... maybe. i haven't tried it yet.
 
  • #22
Stuck again. By using your suggestion, I did
1+(\frac{4}{15}+\frac{15g}{2v_x^2})^2=\frac{10^2}{v_x^2}

v_x^2+v_x^2(\frac{4}{15}+\frac{15g}{2v_x^2})^2=100

v_x^2+v_x^2(\frac{8v_x^2+225g}{(15)(2v_x^2)})^2=100

v_x^2+v_x^2(\frac{8v_x^4+4862025}{900v_x^4})=100

\sqrt{v_x^2+\frac{8v_x^6+4862025v_x^2}{900v_x^4}}=\sqrt{100}

v_x+\frac{2.83v_x^3+2205v_x}{30v_x^2}}=10

30v_x^3+2.83v_x^3+2205v_x=300v_x^2

32.83v_x^3+2205v_x=300v_x^2
Now I am stuck, I don't know how to deal with cubic equation.
 
  • #23
I see that in going from your 3rd step to your 4th step, you kind of 'distributed' the power of 2 in the numerator of the fraction. Whenever you have terms that are being added/subtracted together, like 2 + 3, we can't 'distribute' a power across the terms, for example,
(2+3)^2\neq2^2+3^2
So in the numerator of the fraction in the 3rd step, we unfortunately can not say that
(8v_x^2+225g)^2=8v_x^4+4862025
However, in the denominator of this fraction, your math is correct because you can distribute a power across terms that are being multiplied/divided, for example
(2*3)^2=2^2*3^2

That being said, i'd like to clarify the statement of the problem again, because the numbers don't seem to be working out: You have a projectile that passes between points (0,0,6) and (15,0,10), and the initial velocity is 10 m/s. You have to find the angle that its launched at, right?
 
  • #24
That is correct, and it is driving me insane :)
 
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  • #25
Based on that information, i don't think the problem is physically possible :( the initial velocity isn't enough for the projectile to reach (15,0,10). I think that's what it comes down to. Are you sure the given initial velocity isn't 10m/s in the x or z direction?
 
  • #26
No, it is somewhere in between
 
  • #27
Ok, I did from fourth step
v_x^2+v_x^2\frac{(8v_x^2+2205)^2}{900v_x^4}=100

Where do I go from here? I can't sqrt because sprt(x+y) is not sqrt(x)+sqrt(y), and I don't know if other operations are legal or will get me anywhere. Maybe you are right, it has no answer.
 
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  • #28
If I do it in the first method I did, I get
(\frac{15g}{v+\sqrt{v_z^2-8g}})^2+v_z^2=10^2

\frac{225g^2}{v_z^2+2v_z\sqrt{v_z^2-8g}+v_z^2-8g}+v_z^2=100

(v_z^4+2v_z^3\sqrt{v_z^2-78.4}+v_z^4-78.4v_z^2)+225g^2=100v_z^2+200v_z\sqrt{v_z^2-8g}+100v_z^2-800g

v_z^4+2v_z^3\sqrt{v_z^2-78.4}+v_z^4-78.4v_z^2=100v_z^2+100v_z^2+200v_z\sqrt{v_z^2-78.4}-7840-21609

v_z^4+2v_z^3\sqrt{v_z^2-78.4}+v_z^4-78.4v_z^2=200v_z^2+200v_z\sqrt{v_z^2-78.4}-7840-21609

2v_z^4+2v_z^3\sqrt{v_z^2-78.4}=278.4v_z^2+200v_z\sqrt{v_z^2-78.4}-29449

And I still get stuck. Someone help me, or I will go crazy!
 
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  • #29
Made a mistake above, replace 8g to 8g^2 to all problems, -78.4 to -768.32, 278.4 to 968.32, and -29449 to -98441
 
  • #30
Eerm... Can anyone help me, one last time? Or at least tell me it is unsolvable?
 
  • #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.
 
  • #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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