- #1

- 85

- 2

[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].