- #1
gamesguru
- 84
- 2
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}}.
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}}.
