1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding velocity, then angle

  1. Jun 27, 2008 #1

    IBY

    User Avatar

    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,
     
  2. jcsd
  3. Jun 27, 2008 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to PF IBY,
    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 LaTeX.
     
  4. Jun 27, 2008 #3

    IBY

    User Avatar

    Okay, this thing is not working as I want it to be, I will have to rework it.
     
  5. Jun 27, 2008 #4
    Unclear.
     
  6. Jun 27, 2008 #5

    IBY

    User Avatar

    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.
     
    Last edited: Jun 27, 2008
  7. Jun 27, 2008 #6
    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.
     
  8. Jun 30, 2008 #7

    IBY

    User Avatar

    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.
     
  9. Jun 30, 2008 #8
    You must also multiply the bottom term by -V2z.(fourth step)
     
  10. Jun 30, 2008 #9

    IBY

    User Avatar

    How come?
     
  11. Jun 30, 2008 #10
    Algebraic principle.....

    ex. 100 - 5/5 = 99

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


    100 - 5/5(5) = 99(5) Incorrect
     
  12. Jun 30, 2008 #11

    IBY

    User Avatar

    Ach!*smacked my head* Of course. :)
     
  13. Jun 30, 2008 #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?
     
  14. Jun 30, 2008 #13

    IBY

    User Avatar

    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?
     
    Last edited: Jun 30, 2008
  15. Jun 30, 2008 #14
    Is distance in ft. or meters?

    Also, there must be a time interval or a given angle.
     
  16. Jun 30, 2008 #15

    IBY

    User Avatar

    Meters, sorry. woops
     
  17. Jun 30, 2008 #16
    (a + b)^2 = a^2 + 2ab + b^2. In this case, a is v_z, and b is the square root
     
  18. Jun 30, 2008 #17
    To find the initial velocity we must be given a time interval or launch angle.
     
  19. Jun 30, 2008 #18

    IBY

    User Avatar

    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.
     
    Last edited: Jun 30, 2008
  20. Jun 30, 2008 #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
     
  21. Jun 30, 2008 #20

    IBY

    User Avatar

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Finding velocity, then angle
Loading...