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Vertical trajectory with air resistance

  1. Apr 14, 2008 #1
    1. The problem statement, all variables and given/known data
    a particle P of mass m is projected vertically upwards with velocity [tex]U_0\mathbf{k}[/tex] in a uniform gravitational field [tex]-g\mathbf{k}[/tex] with air resistance proportional to the square of the velocity.

    This is a five part question for which I have found the solutions but there is one method I tried for the last part which lead me to trouble and would like some help clearing it up:

    v) Show that the speed of the particle as i returns to its starting point is:
    [tex] \frac{U_0^2V_t^2}{\sqrt{U_0^2+V_t^2}}[/tex]

    2. Relevant equations
    The equation required to solve (v) is [tex]\frac{dv}{dt}\mathbf{k} = \gamma v^2\mathbf{k} - g\mathbf{k}[/tex] where [tex] \gamma > 0 [/tex] and is a constant of proportionality. (it is a lumped constant made up of [tex] \mu/m[/tex] where [tex]\mu[/tex] is also a constant of proportionality > 0.
    [tex] V_t = [/tex] terminal velocity
    The value of [tex] \gamma[/tex] (not given), I found to be [tex] \frac{g}{V_t^2} [/tex] since the forces acting on P cancel.

    The maximum height reached by P is
    [tex] \frac{V_t^2}{2g}log(\frac{U_0^2+V_t^2}{V_t^2})[/tex] This found by solving the ODE where the sign of [tex]\gamma[/tex] is negative

    3. The attempt at a solution
    There is one step that is giving me trouble but here goes:
    [tex]\frac{dv}{dt}\mathbf{k} = v\frac{dv}{ds}[/tex] (where s(t) is the position of P) such that
    [tex]v\frac{dv}{ds}\mathbf{k} = \gamma v^2\mathbf{k} - g\mathbf{k}[/tex]
    [tex]v\frac{dv}{ds}\mathbf{k} = \frac{g}{V_t^2}( v^2 - V_t^2)\mathbf{k}[/tex] subbing in for [tex]\gamma[/tex]
    [tex]\frac{V_t^2}{g}(\frac{v}{v^2 - V_t^2})\frac{dv}{ds}\mathbf{k} = \mathbf{k}[/tex]
    [tex]\frac{V_t^2}{2g}log|v^2 - V_t^2|\mathbf{k} = (s + C)\mathbf{k}[/tex]

    now at t = 0 (ie: just as P has reached a maximum) s = [tex] \frac{V_t^2}{2g}log(\frac{U_0^2+V_t^2}{V_t^2})[/tex] with v = 0 and so subbing in:
    [tex]C\mathbf{k} = \frac{V_t^2}{2g}(log|- V_t^2| - log(\frac{U_0^2+V_t^2}{V_t^2}))\mathbf{k} [/tex] such that:

    [tex]\frac{V_t^2}{2g}log|v^2 - V_t^2|\mathbf{k} = (s + \frac{V_t^2}{2g}(log\frac{|- V_t^2|V_t^2}{U_0^2+V_t^2}))\mathbf{k}[/tex]

    I now want to solve for v when s = 0
    [tex]|v^2 - V_t^2|\mathbf{k} = \frac{|- V_t^2|V_t^2}{U_0^2+V_t^2}\mathbf{k}[/tex] removed the common factor and exponentiated both sides...Now it's the negative terms in the modulus signs that are causing me bother. If I square and then take the roots to get rid of them I am just as entitled to take the positive square root as I am the negative, only the former leads to:
    [tex]|v| = \frac{U_0^2V_t^2 + 2V_t^4}{\sqrt{U_0^2+V_t^2}}[/tex] which is what I don't want! I(and if I was ignorant as to what my solution should be might be tempted to accept it along with the other...causing much head scratching)

    Am I justified in just saying that I'll take the negative root or is this a poor step?

    For what its worth I reached the correct solution by pulling a minus sign out of the LHS before integrating but no matter how I do it, shouldn't I still reach a solution that is equally true regardless of my method?
  2. jcsd
  3. Apr 14, 2008 #2
    have run out of editting time but the answer to find should have been
    [tex] \frac{U_0V_t}{U_0^2+V_t^2}[/tex] and the problematic answer I get at the end should read as [tex] \sqrt{\frac{U_0^2V_t^2 + 2V_t^4}{U_0^2+V_t^2}[/tex]
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