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Projectile motion with air resistance

  1. Jul 31, 2014 #1
    1. The problem statement, all variables and given/known data

    Model the trajectory of a projectile with an initial velocity [itex]v_0[/itex] that is subject to the force of drag.


    2. Relevant equations

    [itex]F_{\text{d}} = \frac{1}{2}C_{\text{d}}\rho\text{A}(v(t))^2 [/itex]

    [itex]v_x(t) = v_{0x} + \int_{t_0}^t a_x(t')\mathrm{d}t'[/itex]

    3. The attempt at a solution

    I really have no clue how to tackle this problem. I'm sort of leaning towards the idea of a differential equation solution, however I only have a small amount of DE's under my belt from calc 2.

    I have done a fair amount of searching online, however every page that I've seen about this problem assumes drag force is linearly dependent on speed, or even assuming drag is constant. Would I gain enough insight solving the problem in which drag force is linear to make headway on the real life scenario?
     
  2. jcsd
  3. Jul 31, 2014 #2

    mfb

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    If you replace F in your first equation with one of Newton's laws, you get a differential equation with the velocity and its time-derivative.
     
  4. Jul 31, 2014 #3

    rcgldr

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    You need to express acceleration as an equation. mfb gave you a hint for how to do this.

    Then note that acceleration = dv/dt, but in this case, acceleration will be a function of velocity:

    dv/dt = a(v)

    to deal with this rearrange it to:

    dv/a(v) = dt

    then integrate both sides to get velocity versus time (don't forget to include a constant of integration after integrating). You'll need to repeat this process again and integrate to get position versus time.
     
    Last edited: Jul 31, 2014
  5. Jul 31, 2014 #4
    Thanks a lot, both of you. It's been a while since I've done DE's, so I didn't see what now seems so obvious!

    So I get,

    [itex] F_{\text{d}} = -\frac{1}{2}C_{\text{d}}\rho \text{A}v^2 = m\frac{\text{d}v}{\text{d}t} [/itex]

    which then leads me to,

    $$ v(t) = \frac{2m}{K_1 + C_{\text{d}}\rho \text{A}t}$$,

    where [itex] K_1 = C_{\text{d}}\rho \text{A}mK_0 [/itex] and [itex] K_0 [/itex] is just the constant of integration.
     
    Last edited: Aug 1, 2014
  6. Jul 31, 2014 #5
    when I use $$ x(t) = x_0 + \int_{t_0}^{t_1} \frac{2mv_{0x}}{C_d \rho A v_{0x}t + 2m}dt $$

    to get my displacement vs time graph I get $$x(t) = \frac{2m}{C_d \rho A}\ln\left(\frac{2m+C_d\rho A v_{0x}t}{2m}\right) $$

    but this reduces to 0, not [itex] x(t) = v_{0x}t [/itex] if there is no drag. So is this correct, or have I made a mistake?

    [edit] Sorry, I neglected to take a limit as c, p, or A went to zero
     
    Last edited: Jul 31, 2014
  7. Aug 1, 2014 #6

    Here's what I did:

    let [itex] \mu = C_d \rho A [/itex] so then [itex]-\frac{1}{2}\mu v^2 = m\frac{dv}{dt} \implies v\mu t = 2m - vK [/itex].

    so [itex]v = \frac{2m}{\mu t + K} [/itex], where [itex] v(0) = v_{0x} \implies K = \frac{2m}{v_{0x}} [/itex] so then I end up with $$ v = \frac{2mv_{0x}}{2m+\mu v_{0x}t} = \frac{2mv_{0x}}{2m + C_d \rho A v_{0x}t}$$

    which still reduces to the correct equations when taking limits, and we dont have a nasty asymptote.
     
  8. Aug 1, 2014 #7

    rcgldr

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    Sorry for the previous post, I wasn't paying attention to the fact that the force opposes velocity and should be negative: m dv/dt = -1/2 u v^2 . What you have looks correct.

    If the drag is zero, then the integral using CdpA to produce x(t) is undefined as you end up with something in the form of (1/0) ln(1) => 0/0. So if CdpA == 0, the integral becomes:

    $$ x(t) = x_0 + \int_{t_0}^{t_1} \frac{2mv_0}{0 + 2m}dt $$
    $$ x(t) = x_0 + \int_{t_0}^{t_1} v_0dt $$
     
    Last edited: Aug 1, 2014
  9. Aug 1, 2014 #8
    No worries, I actually forgot to add the negative sign in the first occurrence of the force equation, so i made the same mistake :). And in regards to the (1/0)ln(...), it ends up as 0/0 so just use L'Hopital's Rule and you still end up with x(t) = (v_0x)t, same goes for m -> infinity
     
    Last edited: Aug 1, 2014
  10. Aug 1, 2014 #9
  11. Aug 1, 2014 #10

    rcgldr

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    So you enhanced the equations to a 2d situation? I'm not sure I understand the graphs. I assume the graph shows the X Y plane with gravity exerting a force in the -Y direction and drag always opposing velocity (both x and y components), however the peak height reached in the graphs for the object remains the same regardless of drag. It appears the calculator adjusts the initial angle of velocity to do this, but it would seem that if drag is sufficiently large, the peak would be less than the zero drag case, even if the initial velocity was vertical.
     
  12. Aug 1, 2014 #11
    The maximum height reached would not change given any air drag, so long as the drag force is purely horizontal, which is one condition of my problem, although I neglected to mention that. And ya the graphs are a trajectory, horizontal vs vertical distance.
     
  13. Aug 1, 2014 #12

    rcgldr

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    Ok, so an imaginary case where drag is only horizontal. I was expecting something like this:

    http://demonstrations.wolfram.com/ProjectileWithAirDrag

    It might be interesting to plot a postion versus time graph (straight line for zero drag non-zero velocity).
     
    Last edited: Aug 1, 2014
  14. Aug 1, 2014 #13
    Yes. I wanted to originally have drag act as it really would, but I have no clue how to solve a differential equation of the form [itex] q(x) = \frac{dy}{dx} + p(x)y^2 [/itex].
     
  15. Aug 1, 2014 #14

    rcgldr

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    That wolfram article is using an imaginary drag factor that varies with velocity as opposed to velocity squared.

    I though about parametric approach, but apparently the equations involve v and vx and vy:

    $$ m\ \frac{dv_x}{dt} = - \ c\ v \ {v_x} $$
    $$ m\ \frac{dv_y}{dt} = -m\ g \ - \ c\ v \ {v_y} $$

    based on this pdf article that uses numerical integration to solve the problem:

    topic01.pdf

    If the object is falling vertically, then there's no x component, and the equations can be solved:

    wiki free fall with air resistance .htm
     
    Last edited: Aug 1, 2014
  16. Aug 2, 2014 #15

    mfb

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    Do it numerically.
     
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