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Linear Air Resistance need help with a proof

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Equation 2.33 gives the velocity of an object dropped from rest. At first, why v[itex]_{y}[/itex] is small, air resistance should be unimportant and 2.33 should agree with the elementary result V[itex]_{y}[/itex] = gt for free fall in a vacuum. Prove that this is the case. [HINT: remember the Taylor series].


    2. Relevant equations

    equation 2.33---> v[itex]_{y}[/itex](t) = v[itex]_{ter}[/itex](1-e[itex]^{-t/\tau}[/itex])

    v[itex]_{ter}[/itex]=mg/b = g[itex]\tau[/itex]

    3. The attempt at a solution

    I substituted g\[itex]\tau[/itex] for v[itex]_{ter}[/itex] uhhhh and yeah... not sure what to do? I know it's probably really simple but I'm stuck
     
  2. jcsd
  3. Jan 24, 2012 #2

    Curious3141

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    Do you know the Taylor series for ex? Just put x = -t/τ and expand the exponential term to the first few terms. Terms with the power of t greater than one (i.e. t2 and higher) can be disregarded because we're talking about a small time t (after release) here.

    Once you do the simple algebra, you'll get the expression you need.
     
  4. Jan 24, 2012 #3
    yeah the taylor series is e^x = 1 + x

    so I just make it e^(-t/tau) = 1 - (t/tau) ? then do i substitute for tau? sorry I'm just not seeing v = gt :/

    wait so then you plug in 1- (t/tau) into the original equation and then solve it right? haha sorry. I think i figured it out. thank you :)
     
  5. Jan 24, 2012 #4

    Curious3141

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    Just a couple of things: [itex]e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...[/itex] (an infinite series).

    [itex]e^x \approx 1 + x[/itex] is an approximation that's only valid for small x. So you can use it here. However, it's wrong to simply state [itex]e^x = 1 + x[/itex] like you did. Remember that this is only an approximation where you're ignoring all those higher powers of x (because they're too tiny).

    Also, you're not solving an equation, just simplifying an expression. But yes, put (1-t/τ) into that expression in place of [itex]e^{-\frac{t}{\tau}}[/itex], put [itex]v_{ter}[/itex] as gτ, and you'll quickly get the result. :smile:
     
  6. Jan 24, 2012 #5
    haha yeah it was way easier than I was trying to make it :p thank you so much!! :)
     
  7. Jan 24, 2012 #6

    Curious3141

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    No worries! :biggrin:
     
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