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Horizontal and vertical components.

  1. Oct 18, 2009 #1
    I am doing physics homework and I came upon a problem that I had forgotten how to do, and i do not believe my answer is correct. I have tried to Google this but failed, can you please tell me if i have done this correct?
    the question:
    "4. Give the horizontal and vertical components of:
    a.70m/s West
    b. 42m North"

    At first I was wondering how do i solve the horizontal and vertical components of a single variable? So i combined them and hoped they were the same problem.
    203px-DiagramaPotenciasWPde.jpg
    were the 42m is at the right and 70m/s is the base.
    I then found the unknown angle showed in the picture (ignore all the symbols up there)

    [tex]\phi[/tex]=arctan[tex]\frac{42}{70}[/tex]

    [tex]\phi[/tex]=80.54[tex]\circ[/tex]

    and I found the length of the hypotenuses.

    hy=[tex]\sqrt{(l1)^2+(l2)^2}[/tex]

    hy=[tex]\sqrt{(42)^2+(70)^2}[/tex]

    hy=[tex]\sqrt{6664}[/tex]

    hy=81.6

    I then graphed the hypotenuse on a Cartesian Coordinate graph were x= east -x= west
    y=north and -y=south
    and drew my line 81m at 80.54[tex]\circ[/tex] east of north

    and then found my vertical component.

    [tex]\frac{sin\phi}{1}[/tex]=[tex]\frac{op}{hy}[/tex]

    op(1)=sin[tex]\phi[/tex]hy

    op=sin80.54[tex]\circ[/tex]81
    op=79.898
    vertical component= 79.898

    and next my horizontal component.
    hy=[tex]\sqrt{(l1)^2+(l2)^2}[/tex]

    [tex]hy^{2}[/tex]=[tex]l1^{2}[/tex]+[tex]l2^{2}[/tex]

    [tex]hy^{2}[/tex]-[tex]Vc^{2}[/tex]=[tex]Hc^{2}[/tex]

    [tex]81^{2}[/tex]-[tex]79.898^{2}[/tex]=[tex]Hc^{2}[/tex]

    Hc=[tex]\sqrt{81^2-79.898^2}[/tex]

    Hc=13.4

    If i have done this wrong please explain to me how, If i have done them right please explain to me which units I should use for the Hc and Vc and why.
     
  2. jcsd
  3. Oct 18, 2009 #2

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The question, as you have posed it, makes no sense.

    Even though it makes no sense, I'll play ball. If we assume, as you did in your solution, that the north-south direction is "vertical" and the east-west direction is "horizontal", then you already HAVE the horizontal and vertical components of the velocity, so why would you need to try and calculate them?

    If you calculated the hypotenuse and then used trig to calculate the horizontal and vertical components, *then you should have gotten back what you started with.*

    The reason you didn't get back what you started with is because your computation of the angle was wrong.

    arctan(42 / 70) = 30.9637565 degrees
     
  4. Oct 18, 2009 #3
    :frown: It makes perfect sense
    thank you.
     
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