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Homework Help: Air resistance & Trajectory

  1. Aug 17, 2005 #1
    Mathematically, how would air resistance affect trajectory?
    Suppose I shoot a cannonball with initial velocity [itex] v_0 [/itex] at an angle of elevation [tex] \theta [/tex]. Air resistance is a force antiparallel to velocity, represented by the equation (according to Barron's Physics C review book!) [itex] \vec F_{air} = - c \cdot \vec v [/itex], where [itex] c [/itex] is a constant SI-expressed in kg/sec.
    Because the direction of velocity changes with [itex] t [/itex], so does, as well, the direction of air resistance. To represent its effect in the x & y directions, let [itex] \alpha [/itex] represent the instantaneous angle that [itex] \vec v [/itex] makes with the x-axis. Thus, using 2D Cartesian, I represent the velocities as
    [tex] \left\{ \begin{gathered} \vec v_x = \left| {v_0 } \right|\cos \theta - t\left( {c \cdot \vec v} \right)\cos \alpha \hfill \\
    \vec v_y = \left| {v_0 } \right|\sin \theta - t\left( {\vec g + c \cdot \vec v \cdot \sin \alpha } \right) \hfill \\ \end{gathered} \right\} [/tex]
    *But how do I find y(t) and x(t) ? :redface: ?
     
    Last edited: Aug 17, 2005
  2. jcsd
  3. Aug 17, 2005 #2

    Tide

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    I think the equations you are looking for are:

    [tex]\frac {dv_x}{dt} = -c v_x[/tex]

    [tex]\frac {dv_y}{dt} = -g - c v_y[/tex]

    Try solving them for [itex]v_x[/itex] and [itex]v_y[/itex] and then integrating to find x and y.
     
  4. Aug 17, 2005 #3

    Tide

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    Try letting [itex]w = v_y + \frac {mg}{c}[/itex]. That should simplify things. Don't forget to transform back to [itex]v_y[/itex] when you're done.
     
  5. Aug 18, 2005 #4
    (Hehe-this got moved to homework "College level" yet this is not homework and I'm just a curious HS student:smile:)
    Anyway, both differentials are simple enough, so considering mass properly (b/c initially I considered it improperly):
    *Since [itex] v_{x,0} = v_0 \cos \theta [/itex],
    [tex] m\frac{{dv_x }}{{dt}} = - cv_x \Rightarrow v_x = v_0 e^{ - ct/m} \cos \theta [/tex]
    *Since [itex] v_{y,0} = v_0 \sin \theta [/itex],
    [tex] m\frac{{dv_y }}{{dt}} = - mg - cv_y \Rightarrow v_y = \frac{m}{c}\left[ {\left( {g + cv_0 m^{ - 1} \sin \theta } \right)e^{ - ct/m} - g} \right] [/tex]
    -----------------------------
    Hmm...will trajectories affected by air resistance in the form [itex] \vec F_{air} = c \cdot \vec v [/itex] deviate from a parabolic shape? I integrated both [itex] v_x [/itex] and [itex] v_y [/itex] (where [itex] c \ne 0 [/itex]) with respect to time...and it appears that the trajectory deviates from a parabolic shape...hmm..
     
    Last edited: Aug 18, 2005
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