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Trajectory of projectile with friction in air

  1. Jun 4, 2007 #1
    1. The problem statement, all variables and given/known data
    This is not a hw prob, its just something I was trying out.

    To find the equation of the path of the projectile when air friction is considered.


    2. Relevant equations


    F=-bv, where v is a vector.

    3. The attempt at a solution
    Let the projectile be launched with a velocity [tex]v=v_{0x}i+ v_{0y}j+v_{0z}k[/tex];

    Velocity after a time t is: [tex] v(t)=\frac{mv_{0x}}{m+bt}i + \frac{(v_{0y}-gt)m}{m+bt}j + \frac{mv_{0z}}{m+bt}k[/tex]. Integration of this expression wrt t, would give you the path of the projectile... right?
     
  2. jcsd
  3. Jun 4, 2007 #2
    i don't know how you got your function but i know mass usually cancels out

    if you treat friction as a constant force, independent of velocity and time, though i'm pretty friction force in a fluid is not independent of velocity, you just get the regular position function in each component with an acceleration because

    [tex]a_x,z(t)=-k [/tex] and in the x direction you just get a bigger acceleration [tex] a_y(t)=-k+-g [/tex] k and g being both constants you would just get another constant acceleration, bigger.
     
    Last edited: Jun 4, 2007
  4. Jun 4, 2007 #3

    andrevdh

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    I think at higher speeds it is proportional to the square of the speed.
     
  5. Jun 4, 2007 #4
    im bad at this so i would like to test my grasp as well

    [tex]

    a_x(t)=k(\frac{dx}{dt})^2[/tex]
    [tex]
    \frac{d^2(x)}{dt^2}=k(\frac{dx}{dt})^2

    [/tex]

    ?? would this be the correct equation for acceleration for the x component?
    i will try solving the DE if this is correct
     
    Last edited: Jun 4, 2007
  6. Jun 5, 2007 #5
    Err... ok, I took friction to be -bv (vector), so the acceleration in the x and z directions worked out to be [tex]\frac{-b}{m}v_z[/tex]. Since the velocity would change with time as:

    [tex]v_z=v_{0z}-\frac{b}{m}v_zt[/tex], this would give [tex]v_z[/tex] to be:

    [tex]\frac{mv_{0z}}{m+bt}[/tex], where the mass does cancel out dimensionally. Integration of this expression gives a logarithmic function... so, is this right?
     
  7. Jun 5, 2007 #6

    andrevdh

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    The air resistance, [tex]R[/tex], opposes the velocity of the projectile.

    The angle of attack, [tex]\theta[/tex], of the projectile changes in its trajectory.

    By resolving in the x - and y directions we get that

    [tex]m\ddot{x} = - R \cos(\theta)[/tex]

    and

    [tex]m\ddot{y} = - R \sin(\theta) - mg[/tex]

    we can rewrite the sine and cosine terms as

    [tex]\cos(\theta) = \frac{\dot{x}}{v}[/tex]

    and

    [tex]\sin(\theta) = \frac{\dot{y}}{v}[/tex]

    leading to

    [tex]\ddot{x} = -a\dot{x}v}[/tex]

    and

    [tex]\ddot{y} = -a\dot{y}v} - g[/tex]

    An "easier" approach is to rather resolve along the tangential and normal directions to the trajectory - which leads to the equation of the hodograph.
     
    Last edited: Jun 5, 2007
  8. Jun 5, 2007 #7
    Yeah, thats exactly what I've done... just used pure vectors in 3D instead of resolving the force...
     
  9. Jun 6, 2007 #8

    andrevdh

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    I've burned myself once before with this 3D vector stuff, so I would rather keep to what I know - resolving the force components and setting up the equations of motion.
     
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