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Approximating ballistic motion.

  1. Nov 2, 2004 #1
    Hi!

    I'm slightly stuck with an approximation of tmax, the time at which a ballistic projectile reaches maximum height. (projectile shot straigh upwards, no x motion)

    So, by integrating the motion equation

    m * dv/dt = -mg -bv

    (where b is an air drag constant) I've created:

    v(t) = (u + mg/b)*e^(-bt/m) - mg/b

    which at tmax (t at maximum height) must be zero, which then gives:

    tmax = (-m/b) ln [ (mg/b) / (mg/b + u) ]

    where u is the initial velocity in y

    Now when m >> b*tmax, tmax = u/g , howver, I can't get that!

    all I get is (using e^x = 1 + x + x^2/2! +...):

    tmax = u / (mgb/bm + ub/m)

    thanks for any help!

    -tschew
     
  2. jcsd
  3. Nov 3, 2004 #2

    ehild

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    Homework Helper


    You can rewrite your equation labelled with "*" as

    [tex] \exp(\frac{t_{max}b}{m})= 1+\frac{ub}{mg}[/tex]

    As the exponent is much less than 1, you can apply the series expansion and stop at the linear term.

    [tex] \exp(\frac{t_{max}b}{m})=1+\frac{t_{max}b}{m}[/tex]

    [tex] 1+\frac{t_{max}b}{m} = 1 + \frac {ub}{mg}\rightarrow t_{max}=u/g
    [/tex]

    ehild
     
  4. Nov 5, 2004 #3
    Thank you.

    Thanks for the reply. I noticed the minus sign in the ln() expression myself yesterday and got the result :eek:)

    -Tschew
     
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