Max time to reach apex.

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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:
[tex]t_{max}=\frac{\pi}{2}\sqrt{\frac{m}{gk}}[/tex].
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,
[tex]F=ma=-(mg+kv^2).[/tex].
Simplifying,
[tex]v'=\frac{dv}{dt}=-(g+cv^2)[/tex].
Where [itex]c=k/m[/itex].
Separating variables,
[tex]\frac{dv}{g+cv^2}=-dt[/tex].
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:
[tex]-t=\frac{\tan^{-1}v_f\sqrt{\frac{c}{g}}-\tan^{-1}v_0\sqrt{\frac{c}{g}}}{\sqrt{cg}}[/tex].
At the apex,[itex]v=0[/itex], and since [itex]\tan^{-1}0=0[/itex], we find,
[tex]t_{max}=\frac{\tan^{-1}v_0\sqrt{\frac{c}{g}}}{\sqrt{cg}}[/tex]
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].
 

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  • #2
Danger
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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.
 
  • #3
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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.
"(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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