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Projectile Motion problem involving air resistance

  1. Jan 25, 2017 #1
    1. The problem statement, all variables and given/known data
    Ok, so I am attempting to solve a projectile motion problem involving air resistance that requires me to find the total x-distance the projectile traverses before landing again.

    Given:
    [tex]
    \\
    m=0.7\text{kg}
    \\
    k=0.01 \frac{\text{kg}}{\text{m}}
    \\
    \theta=30 \degree
    [/tex]

    2. Relevant equations
    [tex]F_{air}=kv^2[/tex]

    3. The attempt at a solution
    I divided the dynamics of the problem into x-component and y-component equations:
    From [tex]F_{x}=-kv_x^2[/tex], I got:
    [tex]x\left(t\right)=\frac{m}{k}\ln \left(kv_it-m\sec \theta \right)[/tex]

    I divided the y-component of the motion into two parts--going up and going down:
    I used [tex]F_{y1}=-mg-kv_y^2[/tex] for the going up part, and I used [tex]F_{y2}=-mg+kv_y^2[/tex]
    solving the differential equation for the going up part, I got:
    [tex]v_y\left(t\right)=\sqrt{\frac{mg}{k}}\tan \left(\arctan \left(v_{yi}\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{g}{m}}\right)[/tex] -- which is where I am stuck on... because when I plugged in my given values, the graph doesn't look right as its t-intercept is greater than one which is found from solving this kinematically without air resistance.
    solving the differential equation for [tex]v_y(t)[/tex] of the going down part and integrating it, I got:
    [tex]y\left(t\right)=-\frac{m}{k}\ln \left(\cosh \left(t\sqrt{\frac{gk}{m}}\right)\right)+y_i[/tex]

    Can someone help me solve this problem?
     
  2. jcsd
  3. Jan 25, 2017 #2

    ehild

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    No, this is not correct. The force of air resistance is opposite to the velocity vector and its magnitude is proportional to v2. v2 is a scalar, it does not have components.
     
  4. Jan 26, 2017 #3

    Ray Vickson

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    You have not given an initial speed, so your problem is not completely specified.

    For any given initial speed you can set up and solve the DEs numerically, and I think that is about the best you can do (in view of the criticism of "ehild" in #2).

    Perhaps, though, you can get usable approximations by attempting something like a perturbation theory approach in which you essentially expand in powers of the small parameter ##k##.
     
  5. Jan 26, 2017 #4
    Sorry, I forgot to mention that-- the initial velocity is 9 m/s
     
  6. Jan 26, 2017 #5
    What do you mean? Can't any force in vectorspace be divided into
    [PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/18d95a7845e4e16ffb7e18ab37a208d0ab18e0e0, [Broken] https://wikimedia.org/api/rest_v1/media/math/render/svg/3dc8de3d8ea01304329ef9518fad7a6d196c4c01 components?
     
    Last edited by a moderator: May 8, 2017
  7. Jan 26, 2017 #6

    Ray Vickson

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    The friction force acts along the negative of the tangent to the trajectory. See, eg.,
    http://wps.aw.com/wps/media/objects/877/898586/topics/topic01.pdf
    or
    http://young.physics.ucsc.edu/115/range.pdf

    The case of ##\vec{f}_{\text{friction}} = - k \vec{v}## is tractable, but not your case of ##\vec{f}_{\text{friction}} = - k |v|^2\, \vec{v}/|v| = -k |v| \vec{v}##.
     
    Last edited by a moderator: May 8, 2017
  8. Jan 28, 2017 #7
  9. Jan 28, 2017 #8

    Ray Vickson

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    Yes. That is what I said in #3.
     
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