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Homework Help: Help with Projectile Motion problem (including air friction)

  1. Apr 14, 2006 #1
    Hopefully you smart guys can understand this post

    I recently have bought an airsoft pistol that shoots 302fps and I thought it would be fun to calculate the max distance & stuff for the bb including air resistance. The problem is including air resistance into the equation.

    Here's the stats:
    mass of bb= .12g
    diameter of bb= 6mm (.006m)
    muzzle velocity for .12g bb = 302fps (92.0496m/s)

    After reading stuff on the net, I aparently need the equation D=(pCA)/2, where p is the density of air (about 1.2101 kg/m3), C is drag coefficient (around .47 probably?), and A is the area of the bb looking at front (pi*r2 = pi*(.003m)2 = 2.8274*10-5 m2)
    When I plug the numbers in, D= 8.04046*10-6

    I'm sort of lost after this step, aparently Acceleration=-DV2 --> Ax=-DV(Vx) and Ay=-DV(Vy)

    I tried plugging the numbers in but the answer was only somewhere around .04 which I'm pretty sure isn't right at all.

    My main goal is to put it into a parametric function so I can see it visually with distance and time:

    SO, what I'm sure is right (without air resistance):

    What is the air resistance acceleration and how do I plug it into that equation???
  2. jcsd
  3. Apr 14, 2006 #2


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    For small velocities (velocities less than 328 ft/s), air resistance is (approximately) proportional to velocity, rather than the square of velocity.

    Change it to DV from DVĀ² and see if that helps.
  4. Apr 14, 2006 #3
    Changing it to DV instead of DV2 makes the number even smaller (around .00074).

    After reading some more... Acc(drag)=DV2/(mass) and after what you said, should I change that to DV/(mass)? This would make it around 6.168m/s2

    So when I plug it into my equation, is this accurate:

    I know I must be doing something wrong. Please help
  5. Apr 15, 2006 #4


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    I'm sorry, I can't tell which equations of motion you are using. I would recommend using [itex]s = ut + \frac{1}{2}at^2[/itex]. You will need to know the inital height to solve for t, but this is easy to measure.

  6. Apr 15, 2006 #5
    Sorry, thats the equation I was using, I just forgot to put the T in after the first velocities.

    That equation still doesn't include air friction though.
  7. Apr 16, 2006 #6


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    Okay, it seems that you have determined the [negative] acceleration due to air resistance. Now in the x - direction this is the only acceleration experience by the bb round, thus in the x-direction we have;

    [tex]s_{x} = ut + \frac{1}{2}\cdot - a_{drag} t^2[/tex]

    However, in the y - direction, there are two forces acting, gravity and drag but it is important to observe that in this case they are acting in opposite directions. The force of gravity it acting in the same direction as velocity (down) therefore the acceleration will be positive. The drag force is acting in the opposite direction to the velcoity (up). Therefore we can say that;

    [tex]a_{total} = g - a_{drag}[/tex]

    Thus the kinematic equation in the y-direction becomes;

    [tex]s_{y} = ut + \frac{1}{2} (g - a_{drag})t^2[/tex]

    Do you follow?

  8. Apr 16, 2006 #7
    I understand that part. The only problem is that I'm not sure if my adrag calculations are correct.

    Lets work backwards. What's the equation for adrag?
  9. Apr 17, 2006 #8


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    The relationship for drag is given by (as you correctly stated above);

    [tex]F_{drag} = - \frac{1}{2}C\rho Av^2[/tex]

    For a sphere c = 0.5, the density of air is approximately 1.25 kg/m^3.

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