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Projectile motion with wind resistance

  1. Jun 19, 2008 #1
    can anyone explain to me wad is the angle of projection of a projectile motion for the longest range with wind resistance?
  2. jcsd
  3. Jun 19, 2008 #2
    It depends on the initial conditions... namely, the wind resistance.
    You have to have the general equations of motion and optimize distance traveled.
  4. Jun 19, 2008 #3
    i never actually looked into this and calculated it (and am currently too lazy to do it) but wouldnt a 45 degree angle tend to give you the greatest distance... has anyone actually put in the math to see?
  5. Jun 19, 2008 #4
    Yes, a projectile fired at a 45 degree angle from the horizontal will give you the greatest horizontal distance.
  6. Jun 20, 2008 #5


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    Staff: Mentor

    ....but only if there is no wind resistance. With wind resistance, it depends on how much resistance (the more resistance, the lower the angle), as said in post #2.
  7. Jun 21, 2008 #6
    I am going to try to explain this, Please correct me if you see anything wrong with my math or physics, This is just my best guess at this problem.

    First off, to solve with no air resistance (im sure you know this already, but):
    1) find equations that describe the X and Y motion of the projectile
    Code (Text):
    x = Vo*Cos(θ)*t
    y = Vo*Sin(θ)*t + .5*g*t^2
    2) Express time as a function of y displacement:
    Code (Text):
    t = 2*Vo*Sin(θ)/g
    3) Plug in t(θ) into x(t) to find x(θ)
    Code (Text):

    x = 2*Vo^2*Cos(θ)*Sin(θ)/g
    4) Maximize x(θ)
    Code (Text):
    x' = (2 * Vo^2)(cos^2(θ) - Sin^2(θ))
    x' = (2 * Vo^2)(1 - 2 * Sin^2(θ))
    0 = (1 - 2 * Sin^2(θ))
    θ = 45 degrees
    The air resistance complicates things. But I am going to try to use the same method. With air resistance, this will take you up to a certain point where it is algebraically impossible to continue, but conceptually it isnt. Computer aid could be used to find a numerical estimation.

    Ok, here goes:

    The force of the air resistance
    Code (Text):
    F = α * v
    α = a constant, with units Kg/s This depends on the properties of the fluid the projectile it traveling through.
    v = velocity

    So, net acceleration due to air resistance
    Code (Text):
    a =  α * v / m
    a = Β * v
    Β is a constant equal to α / m

    since acceleration isnt constant, in order to write the position equations for the X and Y axis you have to start from the beginning.
    I am going to do X and Y at separate times, as they can get long.


    (acceleration on y axis) = g + B*(velocity on y axis)
    Code (Text):

    d2y/dt2 = g + Β*dy/dt
    d2y/dt2 - Β*dy/dt = g
    Homogeneous solution of the Diffeq.
    Code (Text):
    y = C1 *e^(Βt) + C2*t*e^(Βt)
    for constants C1 and C2.

    Boundary conditions of typical projectile problem are
    Code (Text):
    y(0) = 0
    y'(0) = Vo*Sin(θ)
    Code (Text):

    0 = C1 *e^(Β*0) + C2*0*e^(Β*0)
    C1 = 0
    The solution becomes
    Code (Text):
    y =  C2*t*e^(Βt)
    Code (Text):
    y' =  C2*(t*Β*e^(Β*t) + e^(Βt))
    C2 = Vo*Sin(θ)
    So now we have the homogeneous solution
    Code (Text):

    Yh = Vo*Cos(θ)*t*e^(Β*t)
    Now the particular solution to the Diffeq:
    Code (Text):
    d2y/dt2 - Β*dy/dt = g
    Guess Yp = A*t for some undetermined constant A
    Code (Text):

    Yp = A*t
    Yp' = A
    Yp'' = 0

    d2y/dt2 - Β*dy/dt = g
    0 - Β*A = g
    A = -g/Β
    Yp = -g*t/B
    So now we can write the Y position equation for the projectile with air resistance:
    Code (Text):

    y = Yp + Yh
    [B]y = -g*t/B + Vo*Sin(θ)*t*e^(B*t)[/B]

    Again, we have to start with the acceleration to write the position equation:

    (acceleration on x axis) = -B*(velocity on x axis)
    Code (Text):
    d2x/dt2 = -Β*dx/dt
    Since this one is a homogeneus problem, it is much easier to solve:
    Code (Text):
    d2x/dt2 + Β*dx/dt= 0

    x = C1 *e^(-Βt) + C2*t*e^(-Βt)
    Again, the innitial conditions for projectile motion:
    Code (Text):
    x(0) = 0
    x'(0) = Vo*Cos(θ)
    Plug these in and you get
    Code (Text):
    C1 = 0
    C2 = Vo*Cos(θ)
    So, the final equation for the motion on the X axis is
    Code (Text):

    x = Vo*Cos(θ)*t*e^(-B*t)

    Code (Text):
    [B][SIZE="4"]y = -g*t/B + Vo*Sin(θ)*t*e^(B*t)
    x = Vo*Cos(θ)*t*e^(-B*t)[/SIZE][/B]
    You can then solve for t(θ) from y(t) by setting y=0 (the total y displacement is zero, assuming the projectile lands on the ground).
    Code (Text):
    0 = -g*t/B + Vo*Sin(θ)*t*e^(B*t)
    t = ln(g/(B*Vo*Sin(θ)))/B
    Now plug this into the x(t) equation
    Code (Text):

    x = Vo * Cos(θ) * ln(g/(B*Vo*Sin(θ)))/B * e^(-ln(g/(B*Vo*Sin(θ))))
    And have fun maximizing that...

    I hope this was at least partly right, can anyone see any mistakes or reasons that this is incorrect? There probably are some glaring errors im just not seeing.
    Last edited: Jun 21, 2008
  8. Jun 21, 2008 #7
    And...According to a difference engine:
    Code (Text):
    x' = -((e^-ln[(g Csc[θ])/(B*Vo)]*Vo*Cos[θ]*Cot[θ])/B) + (e^-ln[(g Csc[θ])/(B*Vo)]*Vo*Cos[θ]*Cot[θ]*ln[e]*ln[(g*Csc[θ])/(B*Vo)])/B - (e^-ln[(g Csc[θ])/(B*Vo)]*Vo*ln[(g*Csc[θ])/(B*Vo)]*Sin[θ])/B
    Which reduces to

    Code (Text):
    0 = Cos[θ]*Cot[θ] + Cos[θ]*Cot[θ]*ln[(g*Csc[θ])/(B*Vo)]) - ln[(g*Csc[θ])/(B*Vo)]*Sin[θ]
    Which would give your solution if it were solvable...banking on the slim margain that I was right all this way. It is possible to use a computer, but I doubt I was right all thsi way. Someone look it over pls?
    Last edited: Jun 21, 2008
  9. Jun 21, 2008 #8


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    Homework Helper

    The force of drag is usually approximated as a constant times speed2, but even this is a simplifcation. In real life, the drag force versus velocity is so complicated that tables are used along with interpolation (which isn't linear interpolation unless the table is very dense), especially if speeds transition near or past super-sonic. I seem to recall some ballistics oriented web sites, regarding rifles, and one or more of them had some form of ballistics calculator, but I don't recall if these allowed for very steep angles.

    Another possible source of information would be web sites about motars.
  10. Jun 21, 2008 #9
    I used a constant times velocity, and yes that is a simplification. Air resistance depends on a lot of things, but a constant times velocity is a fairly common simplification.
  11. Jun 21, 2008 #10
    That was a fantastic, thorough, overly kind derivation and explanation swraman!
    I just wanted to add a little*:
    The force of air resistance starts out approximately linearly (as swraman used), and is a good approx for small-medium object at low-medium speeds. For larger objects and higher speeds the resistance goes approximately as a (~)square of the velocity (i totally can't remember the details by i think the exact value of the exponent varies with the shape of the object??), and finally begins to level out asymptotically (this comes from fluid approximations where a fairly large enveloped develops around the projectile shielding it from air resistance).

    *This is just qualitative stuff i remember from an aerodi class, sorry if i'm lacking / messing up any details.
  12. Jun 22, 2008 #11
    Thats interesting, I never knew that :)
    But I havent taken any specific phys classes yet, so I was just gong off of general mechanics.
    I guess its safe to say air resistance is a royal pain to account for, haha.
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