# Max time to reach apex.

1. May 5, 2008

### gamesguru

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

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

2. May 5, 2008

### Danger

I didn't understand any of those characters except for the '=', '+' & '-' signs.
If the thing is launched at greater than escape speed, however, how can there be an apex? It ain't coming back down.

3. May 5, 2008

### gamesguru

"(assuming that $g$ 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.