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Projectile Motion at an angle

  1. Sep 17, 2010 #1
    1. The problem statement, all variables and given/known data
    When the dried-up seed pod of a scotch broom plant bursts open, it shoots out a seed with an initial velocity of 2.64 m/s at an angle of 30.0˚ below the horizontal. The seed pod is 0.465 m above the ground.
    How long does it take for the seed to land?
    What horizontal distance does it cover during its flight?


    2. Relevant equations
    Voy=Vsin[tex]\theta[/tex]
    Vox=Vcos[tex]\theta[/tex]
    h = y0 + v0*t + 1/2 * a * t^2
    d= Vox*t
    3. The attempt at a solution
    I used Voy=2.64*sin30˚, Vox= 2.64cos30 I plugged everything into these equations, solved for t using the quadratic equation, and I got t=0.4723 but that was wrong
     
  2. jcsd
  3. Sep 17, 2010 #2

    rl.bhat

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    Homework Helper

    Show your calculations.
     
  4. Sep 18, 2010 #3
    Voy=2.64sin30= 1.33

    h = y0 + v0*t + 1/2 * a * t^2
    -0.465= 1.33t + 1/2*(-9.8)t^2
    -4.9t^2 + 1.33t+ .465=0
    t= -1.33-sqrt 1.33^2-(4*-4.9*.465)/ 2*-4.9
    t= .04723
     
  5. Sep 18, 2010 #4

    rl.bhat

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    vo, h and g are in the same direction. So they must have the same sign.
     
  6. Sep 18, 2010 #5
    Thank you so much, that helped!
     
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