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Projectile Motion question

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data

    If we include a crude model for the drag force in which the net acceleration on the ball kicked by a football player is given by:
    a = (-k*vx)i + (-g-k*vy)j.
    Derive the equations describing the horizontal and vertical positions as functions of time.
    k=.031 (1/s), vo=69 (ft/s), [tex]\Theta[/tex]0=45.


    2. Relevant equations



    3. The attempt at a solution

    I solved for vx and vy using the information given (vx=v0cos[tex]\Theta[/tex], vy=v0sin[tex]\Theta[/tex] ) plugged these values, along with k, into the acceleration equation. I took the integral of both the horizontal and vertical components independently to get velocity, then integrated that to get the position. The problem is, the horizontal velocity comes out to be -1.51*t, and horizontal position is -.755*t^2, which is obviously wrong because it would be moving backwards the instant it is kicked. What am I doing wrong here? Should I not be solving for vx and vy, and leaving those as variables as well?
     
    Last edited: Feb 10, 2009
  2. jcsd
  3. Feb 10, 2009 #2

    tiny-tim

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    Hi Freyster98! :smile:

    erm :redface: … yes …

    vx and vy are definitely variables …

    the original equation says that the acceleration is -g vertically, and -k (the drag coefficient) times the instantaneous velocity horizontally.
     
  4. Feb 10, 2009 #3
    Im working on the same problem. Do I plug in the initial values and integrate. or do i leave the initial velocity as v0 and then integrate?
     
  5. Feb 11, 2009 #4

    tiny-tim

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    Hi musichael! :smile:

    Leave v0 until the end

    vx and vy are variables …

    integrate, and you will get a constant …

    at that stage you use v0 to find what the constant is. :wink:
     
  6. Feb 11, 2009 #5

    CEL

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    The horizontal velocity is [tex]v_x = v_0 cos \theta_0 + a_x t[/tex], where [tex]a_x = -k v_x[/tex] is a variable and not a constant as you used.
     
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