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Max time to reach apex.

  1. May 5, 2008 #1
    The basic point to this is to show that, no matter how fast an object is thrown up (assuming that [itex]g[/itex] is relatively constant), that there is a maximum time it will take to reach its highest point ([itex]v=0[/itex]), more interesting however is that exact number which strangely involves pi:
    Where [itex]k=\frac{1}{2}\rho A C_d[/itex] is the drag constant.

    We begin by giving the object an upward initial velocity [itex]v_0[/itex].
    Using the drag formula,
    Where [itex]c=k/m[/itex].
    Separating variables,
    The limits can be found by imagining the velocity going from [itex]v_0[/itex] to [itex]v_f[/itex], and time going from [itex]0[/itex] to [itex]t[/itex],
    [tex]\int^{v_f}_{v_0}\frac{dv}{g+cv^2}=-\int^t_0 dt[/tex]
    After some substitution, we arrive, not-surprisingly, at something involving the inverse tangent:
    At the apex,[itex]v=0[/itex], and since [itex]\tan^{-1}0=0[/itex], we find,
    Since it is physically observable that objects thrown upwards more quickly take longer to reach their apex, we take the limit of [itex]t_{max}[/itex] as [itex]v_0\rightarrow\infty[/itex].
    [tex]\lim_{v_0\rightarrow \infty}\frac{\tan^{-1}v_0\sqrt{\frac{c}{g}}}{\sqrt{cg}}=\frac{\pi/2}{\sqrt{cg}}[/tex].
    Substituting back in, we arrive at the final equation for the maximum time it will take an object fired to reach its apex:
    [tex]t_{max}=\frac{\pi}{2}\sqrt{\frac{2m}{g\rho A C_d}}[/tex].
  2. jcsd
  3. May 5, 2008 #2


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    I didn't understand any of those characters except for the '=', '+' & '-' signs. :redface:
    If the thing is launched at greater than escape speed, however, how can there be an apex? It ain't coming back down.
  4. May 5, 2008 #3
    "(assuming that [itex]g[/itex] is relatively constant)"
    This means that g=9.8 m/s^2 and stays sufficiently close to 9.8. But, you're right, if it is launched fast enough, it won't come back down.
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