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Homework Help: Projectile motion with air resistance proportional to velocity

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data
    The projectile with initial velocity [tex]v_0[/tex] is moving in environment with air resistance, which is linear to projectile's velocity. The task is to prove, that horizontal distance is maximal, if the elevation angle ([tex]\alpha[/tex]) and the angle of trajectory tangent in place of impact are complementary.

    2. Relevant equations
    [tex] F_x = - k.v_x [/tex]
    [tex] F_y = - m.g - k.v_y [/tex]
    [tex] v_x = v.\cos(\varphi) [/tex]
    [tex] v_y = v.\sin(\varphi) [/tex]
    [tex] v_{0x} = v_0.\cos(\alpha) [/tex]
    [tex] v_{0y} = v_0.\sin(\alpha) [/tex]

    3. The attempt at a solution
    equations in [tex]x[/tex] direction:
    [tex] a_x = \frac{F_x}{m} = - \frac{k}{m} v_x [/tex]
    [tex] a_x = \frac{\mathrm{d}v_x}{\mathrm{d}t} = - \frac{k}{m} v_x [/tex]
    [tex] \frac{\mathrm{d}v_x}{v_x} = - \frac{k}{m}\mathrm{d}t \Rightarrow v_x = v_{0x}.e^{-\frac{k}{m}t} [/tex]
    after integration:
    [tex] x = -\frac{m}{k} v_{0x} . e^{-\frac{k}{m}t} + \frac{m}{k} v_{0x} [/tex]

    equations in [tex]y[/tex] direction:
    [tex] a_y = \frac{F_y}{m} = - g - \frac{k}{m} v_y [/tex]
    [tex] a_y = \frac{\mathrm{d}v_y}{\mathrm{d}t} = - g - \frac{k}{m} v_y [/tex]
    [tex] \frac{\mathrm{d}v_y}{- g - \frac{k}{m} v_y} = \mathrm{d}t [/tex]
    after integrations and mathematical expressing:
    [tex] v_y = \frac{m}{k}\left(g + \frac{k}{m}v_{0y}\right)e^{-\frac{k}{m}t} - g\frac{m}{k} [/tex]
    [tex] y = - \frac{m^2}{k^2}\left(g+ \frac{k}{m}v_{0y}\right).\left(e^{-\frac{k}{m}t} - 1\right) - g\frac{m}{k}t [/tex]

    And here is my problem:
    I would like to express from [tex]y[/tex] equation the time of the impact [tex]t_i[/tex]. I know about this moment, that [tex] y = 0 [/tex]. If I am able to write explicit function for [tex] t_i(\alpha) [/tex], then I would pass it to [tex] x [/tex] function. Afterwards I would differentiate this [tex] x [/tex] function according to [tex] \alpha [/tex]. Then I would find the ideal [tex] \alpha [/tex], which I would use to find [tex] t_i [/tex] which I would finally use to find impact angle.
    Unfortunatelly, I can't see the way to find explicit expression of [tex] t_i [/tex] from [tex] y = 0 [/tex] function.
     
  2. jcsd
  3. Apr 25, 2010 #2

    gabbagabbahey

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    Basically, you want to maximize [itex]x(t_i)-x(0)[/itex] subject to the constraint [itex]y(t_i)=0[/itex]...that seems like a prime candidate for the mathod of Lagrange multipliers to me:wink:
     
  4. Apr 26, 2010 #3
    Thank you for your answer!
    Basically, you say (and I think you are right...), there are two other possibilities:
    • from function [tex] y = 0 [/tex] not to express [tex] t_i(\alpha) [/tex], but [tex] \alpha(t_i) [/tex] and to use this in function [tex] x(t_i) [/tex], which I should maximize
    • to use method of Lagrange multipliers

    First mentioned method brings me to situation, that I am not able to express [tex] t_i [/tex], which would maximize [tex] x(t_i) [/tex] function, anyway.
    Second, I was thinking about method of Lagrange multipliers before, but I tried to avoid it, because I am trying to help with this homework to my friend, who is first-year student at university (industrial management) and he doesn't know the method yet... But I think, I will have to.
     
  5. Apr 26, 2010 #4

    gabbagabbahey

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    Gold Member

    Hmmm.... if this is supposed to be a 1st year level problem, there may be a simpler method...perhaps energy conservation would make things easier.
     
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