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Another motion prob

  1. Sep 13, 2007 #1
    [tex]_{}[/tex]1. The problem statement, all variables and given/known data

    If a long jumper at the top of his projectory is moving at 6.5 m/s (horizontally) and his cantre of mass is 1.1m above where it was when he launched into the jump, how fast must he have been moving when he launched?

    2. Relevant equations

    v[tex]_{x}[/tex] = v[tex]_{0x}[/tex] + a[tex]_{x}[/tex]t
    x = [tex]\frac{1}{2}[/tex] (v[tex]_{0x}[/tex] + v[tex]_{x}[/tex])t
    x = v[tex]_{0x}[/tex]t + [tex]\frac{1}{2}[/tex]a[tex]_{x}[/tex]t[tex]^{2}[/tex]
    v[tex]_{x}[/tex][tex]^{2}[/tex] = v[tex]_{0x}[/tex][tex]^{2}[/tex] + 2a[tex]_{x}[/tex]x

    3. The attempt at a solution

    v[tex]_{x}[/tex] = 6.5 m/s
    y = 1.1m
    y[tex]_{0}[/tex] = 0m
    while the jumper is at the top of trajectory [tex]\rightarrow[/tex] v[tex]_{y}[/tex] = 0m/s

    is it possible to work out this question with just those equations?
    they all have either a time, acceleration of x component (i.e. distance travelled)
    also is this considered a projectile motion problem as he was running before he jumped.
     
  2. jcsd
  3. Sep 13, 2007 #2

    learningphysics

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    Homework Helper

    HInt: what is the horizontal velocity when he jumped? what is the vertical velocity when he jumped? Find these 2 separately... then you can get the velocity with which he jumped.
     
  4. Sep 13, 2007 #3
    thanks learningphysics

    cool so it's:

    vy[tex]^{2}[/tex] = voy[tex]^{2}[/tex] - 2g[tex]\Delta[/tex]y
    0[tex]^{2}[/tex] = voy[tex]^{2}[/tex] - 2[tex]\times[/tex]9.81[tex]\times[/tex]1.1
    voy[tex]^{2}[/tex] = 21.582
    voy = 4.65 m/s

    vx = vox
    voy = 6.5 m/s

    vo = [tex]\sqrt{voy[tex]^{2}[/tex] + vox[tex]^{2}[/tex]}[/tex]
    = [tex]\sqrt{4.65[tex]^{2}[/tex] + 6.5[tex]^{2}[/tex]}[/tex]
    = [tex]\sqrt{21.582 + 42.25}[/tex]
    = [tex]\sqrt{63.832}[/tex]
    = 7.989 m/s [tex]\rightarrow[/tex] 8 m/s
     
  5. Sep 13, 2007 #4

    learningphysics

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    Looks good!
     
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