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Projectile Motion

  1. Dec 17, 2006 #1
    1. The problem statement, all variables and given/known data
    A star in the long jump goes into the jump at 12 m/s and launches herself at 20.0 degrees above the horizontal. How long is she in the air before returning to Earth? (g=9.81 m/s2)



    2. Relevant equations
    first I look for the v0(cos) and then I look for vfy= v0(sin)
    I used vfy/g and I don't get the answer. I have the answer which is .83 s
    but I can't get it on my own...


    3. The attempt at a solution

    I used vfy/g to get the time, but I don't get the correct answer I get .4
     
  2. jcsd
  3. Dec 17, 2006 #2

    ShawnD

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    Science Advisor

    You probably just forgot the downfall time. Here's what the setup is like for finding the half-time in the air.
    0 = (m)(v) - (F)(t)
    Add a sine function to represent the vertical velocity
    0 = (m)(v)sin(theta) - (m)(g)(t)
    factor out the mass and start filling in values
    0 = (12)sin(20) - (9.8)(t)
    t = 0.4188 going up
    Add another 0.4188 for the going down to get a total of 0.8376 seconds.
     
  4. Dec 17, 2006 #3
    thanks! yes I get it now, I was forgetting to multiply the time with the gravity.
    thanks again. :)
     
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