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Parabolic Air Resistance

  1. Jan 16, 2008 #1
    Hey, I have a question about how to calculate the air resistance, and eventually how much more energy is required to get the same result, once air resistance has been factored in.

    I have thought of two ways to solve the problem: One way requires multi-variable calculus, which I havn't had, the other way entails breaking the equation into x & y components.

    The problem I have is as fallows. I have a tennis ball, which has a mass of 57g, that is fired at velocity V and at angle A. If air resistance can be ignored, it will land right on target, however, air resistance can not be ignored. I need to find how much I need to increase the initial velocity, in order to achieve the same result, or alternatively how much I need to change the launch angle.

    It seems like this would be a common ballistics question, but I have been unable to find an equation. Any help is greatly appreciated! Thanks in advance, - Peter

    P.S. Here is a good calculator that is related to my problem http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html
    Last edited: Jan 16, 2008
  2. jcsd
  3. Jan 16, 2008 #2

    Andy Resnick

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    Part of the problem could be that you are allowing too many things to vary. For example, given a target, it's possible to vary either A or V (or both)- the two are combined into V*cos(A) and V*sin(A).

    In any case, adding air resistance is not too difficult- the equation to solve is:

    m ds/dt - D*s = -mg, where D is the drag coefficient (6*pi*R*viscosity) for low Reynolds numbers. s = dv/dt.

    Neglecting air resistance, the way to solve the problem is:

    m d^2r/dt^2 = -mg, or m ds/dt = -mg which resolves to

    m x = c*t
    m y = -1/2 mgt^2 + bt + d

    putting in V_x(0) = V*cos (A) and V_y(0) = V*sin(A) and x(0) = y(0) = 0 fixes the constants, and the energy put into the ball is 1/2 mV^2. The target point is found by solving for y(t_f) = 0, and x(t_f) is the target.

    So, go through the same procedure starting instead from m ds/dt - D*s = -mg. This time, the position of the target is known, y(0) is also known, but the initial velocities are not known. So, solve for those, and then the input energy is still 1/2 mV^2.
  4. Jan 23, 2008 #3
    Sorry I took so long to get back to you. I've had internet outages for the past week or so. :( I worked over your equations, however I am still a little bit confused. In your first equation m ds/dt - D*s = -mg, where D is drag coefficient (6*pi*R*viscosity), if you could walk me through what each variable represents.

    For my drag coefficent, a tennis ball in this case, my Reynolds number turns out to be 135000. However this yeilds a giant drag coefficient.

    Thanks for the help so far, - Peter
  5. Jan 23, 2008 #4

    Andy Resnick

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    Hi Peter,

    discussing equations is nearly impossible with ASCII, but here goes:

    m ds/dt - D*s = -mg is Newton's first law, s = velocity = dr/dt. I wrote the equation as a linear first-order differential equation becasue it's easier to solve. The solution is easy to find and too messy to try and write down here. The velocity terms will look something like what's on the Wiki page:


    You have different initial conditions, but the exponential component to the velocity is the key effect of air resistance.

    The drag I wrote is for Stokes flow- slow flow. Your Reynolds number is definitely *not* small. If you already have a drag coefficiant, that's great- ignore the Stokes equation I put down.

    Does this help?
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