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Golf Balls and differential equations

  1. Oct 12, 2012 #1
    Hey Everybody, I am supposed to model the trajectory of a golf ball. I have been given the equations for velocity as a function of its derivative with respect to time. I am supposed to find the x-range as a function of the angle θ. (Pardon my bad latex skills, I will fix mistakes):

    1. The problem statement, all variables and given/known data
    These are the equations which have a mathematical solution, and they do not include lift. -.25 is the drag coefficient on the golf ball.


    -.25v[itex]_{x}[/itex] = [itex]\frac{dv_{x}}{dt}[/itex]
    and
    -.25v[itex]_{y}[/itex] -g = [itex]\frac{dv_{y}}{dt}[/itex]

    Therefore

    [itex]\frac{dv_{y}}{dt}[/itex] +.25 v[itex]_{y}[/itex] = -g where g is the earth's acceleration due to gravity.
    and
    [itex]\frac{dv_{x}}{dt}[/itex] +.25v[itex]_{x}[/itex] = 0

    2. Relevant equations

    Integrating factor: e[itex]^{\int P(t) dt}[/itex]
    x range = v[itex]_{i}[/itex]cosθ * t

    3. The attempt at a solution

    For v[itex]_{x}[/itex]:

    I(t) = e[itex]^{\int P(t) dt}[/itex]
    I(t) = e[itex]^{.25t + k_{1}}[/itex]

    [itex]\int(d e^{.25t}e^{k_{1}}v_{x} /dt)[/itex] = [itex]\int 0 dt[/itex]

    e[itex]^{.25t}[/itex]e[itex]^{k_{1}}[/itex]v[itex]_{x}[/itex] = C[itex]_{1}[/itex]

    v[itex]_{x}[/itex] = C[itex]_{1}[/itex]e[itex]^{.25t}[/itex] because e[itex]^{k_{1}}[/itex] is just a constant too.

    v[itex]_{i}[/itex]cos([itex]\Theta[/itex]) = C[itex]_{1}[/itex]e[itex]^{.25t}[/itex]

    I use the statutory initial velocity of a golf ball of 76.2 m/s.

    cos([itex]\Theta[/itex]) = [itex]\frac{C_{1}}{76.2}[/itex]e[itex]^{.25t}[/itex]

    [itex]\Theta[/itex] = cos[itex]^{-1}[/itex]([itex]\frac{C_{1}}{76.2}[/itex]e[itex]^{.25t}[/itex])

    For v[itex]_{y}[/itex]: (skipping the prelim stuff)

    e[itex]^{.25t}[/itex]e[itex]^{k_{2}}[/itex]v[itex]_{y}[/itex] = -gt + C[itex]_{2}[/itex]

    v[itex]_{y}[/itex] = e[itex]^{-.25t}[/itex]e[itex]^{-k_{2}}[/itex](-gt + C[itex]_{2}[/itex])

    v[itex]_{i}[/itex]sin(θ) = e[itex]^{-.25t}[/itex]e[itex]^{-k_{2}}[/itex](-gt + C[itex]_{2}[/itex])

    sin(θ) = [itex]\frac{e^{-.25t}e^{-k_{2}}(-gt + C_{2})}{76.2}[/itex]

    θ = sin[itex]^{-1}[/itex]([itex]\frac{e^{-.25t}e^{-k_{2}}(-gt + C_{2})}{76.2}[/itex])

    These equations for θ seem pretty nasty, not to mention I have no way of knowing the Constants because I only know the absolute value of the velocity, not the components.
    Also, these equations I have found for θ have seemingly nothing to do with range, they are a function of time. Any hints? Should I use another solution method for the v[itex]_{y}[/itex] differential equation? Laplace Transform?
     
  2. jcsd
  3. Oct 12, 2012 #2
    I tried to solve v[itex]_{y}[/itex] using Laplace Transforms.

    I got

    v[itex]_{y}[/itex] = e[itex]^{-.25t}[/itex](v[itex]_{y}[/itex](0) +4g) - 4g

    I still dont know v[itex]_{y}[/itex](0) because the angle can vary.
     
  4. Oct 13, 2012 #3
    Getting closer, can someone tell me if there is a way to write initial velocity in y as a function of θ?
     
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