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Projectile motion given theta and x

  1. Jul 5, 2009 #1
    1. The problem statement, all variables and given/known data

    An Olympic long jumper leaves the ground at an angle of 23o and travels through the air for a horizontal distance of 8.7m before landing. What is the takeoff speed of the jumper?

    2. Relevant equations

    8.7m = cos(23o)Vot

    y = sin(23o)t + (1/2)(-9.8m/s2)t2

    0m2/s2 = sin2(23o)Vo2 + 2(-9.8m/s2)y

    3. The attempt at a solution

    Fifteen minutes and many permutations later, I get t = .835, .782 with Vo = 11.32m/s, 12.09m/s

    The answer in the book is 11m/s.


    Can you explain the steps you would take to solve this? It literally took me 15 minutes of mathematical manipulation to isolate t. The final step to find t for me was:

    .697 = t2(3.6934 - 4.9t2)


    Thanks guys!
     
  2. jcsd
  3. Jul 5, 2009 #2
    y = sin(23)*V0*t + (1/2)(-9.8m/s2)t
    u missed a Vo here.
    well get t from the first equation, and substitute it in t in the second one.
    from the first equation u get
    t=8.7 / cos(23)Vo
     
  4. Jul 5, 2009 #3

    ideasrule

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    An easier way is to solve for t first in y = sin(23)*V0*t + (1/2)(-9.8m/s2)t^2:

    0=sin(23)*V0*t + (1/2)(-9.8m/s2)t^2
    Divide out t...
     
  5. Jul 6, 2009 #4

    andrevdh

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    [tex]x=V_o \cos(\theta_o)t[/tex]

    giving

    [tex]t = x\frac{x}{V_o \cos(\theta_o)}[/tex]

    and from

    [tex]y = V_o \sin(\theta_o)t - 0.5gt^2[/tex]

    we have that

    [tex]y = x\tan(\theta_o) - \frac{gx^2}{2V_o\cos^2(\theta_o)}[/tex]

    the parabolic equation describing the trajectory of the projectile
     
    Last edited: Jul 6, 2009
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