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Homework Help: Projectile motion

  1. Sep 26, 2010 #1
    1. The problem statement, all variables and given/known data

    An object is launched from ground at an angle of 73 degrees from horizontal and reaches a maximum height of 47 meters. What is magnitude of object's velocity when object's horizontal displacement is 23 meters? Answer in m/s. Answer is 10.61.
    [tex]\theta[/tex]=73
    [tex]\Delta[/tex]y=47m
    [tex]\Delta[/tex]x=23m

    2. Relevant equations
    [tex]\Delta[/tex]y=v0y2/2g
    [tex]\Delta[/tex]x=v0xt
    vfx=v0x
    [tex]\Delta[/tex]y=v0yt-1/2gt2
    y=tan[tex]\theta[/tex]x-g/(2v02cos2[tex]\theta[/tex])*x2
    3. The attempt at a solution
    47=v0y2/(2*9.81)
    v0y=[tex]\sqrt{}(2)(9.81)(47)[/tex]=30.37
    30.37=v0sin73
    v0=31.76
    By using last equation, I got y=45.13, y coordinate when horizontal displacement is 23 meters.
    vfy=[tex]\sqrt{}(30.37)2-(2*9.81*45.13)[/tex]=6.07
    t=45.13
    23=v0x*45.13
    v0x=.51
    v=6.09
     
    Last edited: Sep 27, 2010
  2. jcsd
  3. Sep 26, 2010 #2

    Delphi51

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    Homework Helper

    I agree with your calc down to v0=31.76, but don't know where you got "y=45.13" or what it means. I didn't get the specified answer either, but maybe we can pool our work to find it!

    initial horiz velocity = 31.76*cos(73) = 9.28
    Time to x = 23: 23 = 9.28*t, so t = 2.477
    vertical velocity at this time: Vy = Voy + at
    Vy = 30.37-9.81*2.477 = 6.07
    combined speed at time 2.477: v² = 9.28² + 6.07², v = 11.09
     
  4. Sep 26, 2010 #3

    JJBladester

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    Gold Member

    I understand a lot of times working with [tex] tags you may have to edit your post, but each time you do, you will see additional tags as below:

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution


    For some reason the forum adds them again each time you edit your post, so be careful to just remove them.
     
  5. Sep 27, 2010 #4
    Answer calculated is agreeable, but I can't understand what's posted beneath it. Are you willing to explain it to me?
     
  6. Sep 28, 2010 #5

    JJBladester

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    Gold Member

    eestep, please clarify what you are need help with (write it out fully).
     
  7. Sep 30, 2010 #6
    I appreciate all of your help and attempted it again successfully.
     
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