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Projectile motion w/drag equations

  1. May 19, 2007 #1
    Hello. How do we derive:
    Maximum distance of a projectile as a function of initial velocity and launch angle. [taking into account the air resistance] Assuming mass of projectile, air viscosity, drag coefficient, etc are constants.

    Is there any sites giving the steps?
     
  2. jcsd
  3. May 19, 2007 #2
    I have another question: I have researched that
    for horizontal component (x), Fdrag= -0.5 C p A V Vx
    for vertical component (y), Fdrag= -0.5 C p A V Vy
    where V=square root(Vx^2+Vy^2)

    Isn't Fdrag for horizontal component is simply= -0.5 C p A Vx^2
    and for vertical component, Fdrag = -0.5 C p A Vy^2 ?
     
  4. May 19, 2007 #3

    Mentz114

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    The horizontal and vertical drag components will depend on the angle, surely ?
    Can you find the distance flown when there is no drag ?
     
  5. May 19, 2007 #4
    yes i can. I have found the answer to my second post. Any answers on my first post?
     
  6. May 19, 2007 #5

    Mentz114

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    Gold Member

  7. May 19, 2007 #6
    I think you are talking about projectile motion without air resistance?
    Nonetheless, in projectile with air resistance, is it possible to obtain an equation of maximum distance as a function of initial velocity and launch angle (as the variables only)?
     
  8. May 19, 2007 #7

    Mentz114

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    Gold Member

    A good approach would be to write the equation for the maximum distance as a function of V and theta, then add terms to the derivation to account for air resistance. To get the maximum range you work out the time the projectile is in the air from the vertical component, then multiply that time by the horizontal velocity. If there's air resistance, the time in the air becomes a tricky calculation.

    If there's air resistance it will enter the formula.
     
  9. May 19, 2007 #8
    It's not that simple as horizontal velocity is changing due to air drag.
     
  10. May 19, 2007 #9
    And btw why is this thread moved? I really don't think this is introductory physics.
     
  11. May 20, 2007 #10

    Mentz114

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    Who said it was simple ? The drag is obviously going to affect both components. Why don't you have a shot at it ? I'm willing to help if you show some work. You could also try looking for notes on the web.
     
  12. May 21, 2007 #11
    I want to integrate

    dVx/dt = -(D v Vx) / m

    but v=sqrt(Vx^2+Vy^2)
    I have tried substitution method to eliminate Vy, but it's not possible (I have done this for hours).

    My question is,
    Is it even possible to come up with an equation of horizontal distance as a function of launch angle, initial velocity and time?
     
  13. May 21, 2007 #12
    I cannot say for sure no, but I didn't see any such function so far.

    However, I wonder why you need the closed form? If it is a goal unto itself, well... But if you need to compute this as part of some other problem, numeric solution can be easily constructed in many ways. The details are nicely presented in the link given by Mentz114.

    --
    Chusslove Illich (Часлав Илић)
     
  14. May 21, 2007 #13
    Well I have a set of equations, only based on that Fdrag= -0.5 c p a v^2. I just divide into horizontal and vertical component, mainly the force and acceleration. However to find the function of horizontal distance, I believe I have to integrate dVx/dt and then integrate it again.

    While combining the equations (hoping that I can eliminate the unknown variable), I found a strange inconsistency in the equations:

    (A=acceleration)

    Fx= -D v^2 cosX = -D v Vx

    Fy= -D v^2 sinX - mg = -D v Vy -mg

    F=sqrt(Fx^2+Fy^2)
    A=dv/dt= [sqrt(Fx^2+Fy^2)] /m ---------------[1]

    v=sqrt(Vx^2+Vy^2)
    dv/dt= (Vx Ax + Vy Ay)/ sqrt(Vx^2+Vy^2) --------[2]

    Solving [1]=[2], they are not equal and after solving [1]=[2], I found that Vx=0 for [1] and [2] to be equal. Yet Vx is obviously not 0.

    Can anyone explain why? I know they are time-consuming to solve (not difficult though). Thnx.
     
  15. May 21, 2007 #14
    Your expression [2] is not an equation, but an indentity: left hand side is acceleration component in direction of velocity, right hand side is the same, only using scalar product formulation.

    Equation [1] is wrong: left hand side is same as in [2], but right hand side is full acceleration, not only the component in velocity direction. The corrected [1] would be alike [2]:
    [tex]
    \frac{dv}{dt} = \frac{v_x F_x + v_y F_y}{\sqrt{v_x^2 + v_y^2}} \frac{1}{m}
    [/tex]

    But with [2] being identity, [1] and [2] would only assert:
    [tex]
    a_x = \frac{F_x}{m}, \quad a_y = \frac{F_y}{m}
    [/tex]
    and you're back to square one.

    --
    Chusslove Illich (Часлав Илић)
     
  16. May 22, 2007 #15
    What's the difference between identity and equation?
    So it's not possible to integrate and come up with distance equation?
     
  17. May 22, 2007 #16
    (I may not be using proper mathematical terms, so sorry for any confusion.)

    Identity is something that is true by definition, so it cannot provide any links between the independent quantities in a particular problem. Equation is a constraint on the problem imposed from outside, e.g. by physical laws. Thus, your expression [1] comes from 2nd Newton law (an equation), whereas [2] comes from definitions of tangential acceleration and vector product (an identity).

    I cannot claim that it is not possible (math is not my strong point), I just haven't seen any such integration.

    --
    Chusslove Illich (Часлав Илић)
     
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