Finding velocity, then angle

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  • #1
IBY
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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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  • #2
Hootenanny
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Welcome to PF IBY,
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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  • #3
IBY
106
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Okay, this thing is not working as I want it to be, I will have to rework it.
 
  • #5
IBY
106
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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)
[tex]10^2-(\frac{15*9.8}{v_z+\sqrt{v^2_z-8*9.8}})^2=v_z[/tex] (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 [tex]V^2_z=10^2-V^2_x[/tex], but no matter what I do, my results of Vz always ends up in the end making [tex]V^2_z+V^2_x[/tex] 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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  • #6
335
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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.
 
  • #7
IBY
106
0
Ok, but it doesn't work. Look:
[tex]10^2-(\frac{15g}{v_z+\sqrt{v^2_z-8g}})^2-v_z^2=0[/tex]
I squared term x and 10.
[tex]100-\frac{225g^2}{v_z^2+v_z^2-8g}-v_z^2=0[/tex]
Added like terms on the bottom of fraction and moved 100 to other side.
[tex]-\frac{225g^2}{2v_z^2-8g}-v_z^2=-100[/tex]
Multiply the bottom of fraction to both sides.
[tex](2v_z^2-8g)\frac{-225g^2}{2v_z^2-8g}-v_z^2=-100(2v_z^2-8g)[/tex]

[tex]-225g^2-v_z^2=-200v_z^2+800g[/tex]
Transferred two terms and coverted g to 9.8 and solved rhs.
[tex]199v_z^2=800g+225g^2[/tex]

[tex]199v_z^2=800*9.8+225*9.8^2[/tex]

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

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

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

[tex](\frac{147}{12.16+\sqrt{12.16^2-78.4}})^2+12.16^2[/tex]

[tex](\frac{147}{12.16+8.33})^2+147.87[/tex]

[tex](\frac{147}{20.49})^2+147.87[/tex]

[tex]51.47+147.87[/tex]

[tex]\sqrt{199.34}[/tex]
Which is equal to 14.12, which is not equal 10, so, yeah, I am stuck, still.
 
  • #8
21
0
You must also multiply the bottom term by -V2z.(fourth step)
 
  • #9
IBY
106
0
How come?
 
  • #10
21
0
Algebraic principle.....

ex. 100 - 5/5 = 99

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


100 - 5/5(5) = 99(5) Incorrect
 
  • #11
IBY
106
0
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

[tex]
(\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}
[/tex]

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
IBY
106
0
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
[tex]10=-\frac{1}{2}gt+v_zt^2+6[/tex] and [tex]15=v_xt[/tex]
In the second equation, [tex]v_x=\frac{15}{t}[/tex]
In the first one, I moved 10 to the other side, making it
[tex]t=\frac{v_z+\sqrt{v_z^2-8g}}{g}[/tex] (plus in the quadratic cause I want to choose the side when the ball hits the spot) I insert this in [tex]v_x=\frac{15}{t}[/tex], making initial velocity

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

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 [tex]v_x^2 +v_z^2=10^2[/tex], 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
21
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Is distance in ft. or meters?

Also, there must be a time interval or a given angle.
 
  • #15
IBY
106
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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
21
0
To find the initial velocity we must be given a time interval or launch angle.
 
  • #18
IBY
106
0
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
21
0
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
IBY
106
0
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,

[tex]
(1)\:\:\:\:\:4=-\frac{1}{2}gt^2+v_zt
[/tex]

We also have, in the x direction, that
[tex]
(2)\:\:\:\:\:t=\frac{15}{v_x}[/tex]

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

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
[tex]
m=\frac{v_z}{v_x}=tan\theta[/tex]

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

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

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

If we now solve for vx2 and substitute into eqn (5), we can solve for m, and thus find the solution... maybe. i havent tried it yet.
 
  • #22
IBY
106
0
Stuck again. By using your suggestion, I did
[tex]1+(\frac{4}{15}+\frac{15g}{2v_x^2})^2=\frac{10^2}{v_x^2}[/tex]

[tex]v_x^2+v_x^2(\frac{4}{15}+\frac{15g}{2v_x^2})^2=100[/tex]

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

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

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

[tex]v_x+\frac{2.83v_x^3+2205v_x}{30v_x^2}}=10[/tex]

[tex]30v_x^3+2.83v_x^3+2205v_x=300v_x^2[/tex]

[tex]32.83v_x^3+2205v_x=300v_x^2[/tex]
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,
[tex](2+3)^2\neq2^2+3^2[/tex]
So in the numerator of the fraction in the 3rd step, we unfortunately can not say that
[tex](8v_x^2+225g)^2=8v_x^4+4862025[/tex]
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
[tex](2*3)^2=2^2*3^2[/tex]

That being said, i'd like to clarify the statement of the problem again, because the numbers dont 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
IBY
106
0
That is correct, and it is driving me insane :)
 
Last edited:
  • #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?
 

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