Edmund Conti

2021-11-13

To evaluate: The terminal velocity from a graphical analysis.

Ancessitere

Given that a body of mass m is projected vertically downward with initial velocity vo.
It is known that force $F=ma$ and $F={F}_{p}—{F}_{r}$, where ${F}_{p}=mg$ and ${F}_{r}$ is the air resistance.
Given that the resisting force is proportional to $\sqrt{v}$.
That is, ${F}_{r}\propto \sqrt{v}$ which implies ${F}_{r}=k\sqrt{v}$.
Substitute the values ${F}_{p}=mg,{F}_{r}=k\sqrt{v}$ in the expressions in $F={F}_{p}-{F}_{r}$ and evaluate as follows.
$F={F}_{p}-{F}_{r}$
$ma=mg-k\sqrt{v}$
$a=g-\frac{k}{m}\sqrt{v}$
$\frac{dv}{dt}=g-\frac{k}{m}\sqrt{v}\left[\begin{array}{c}\because a=\frac{dv}{dt}\end{array}\right]$
The value of v is evaluated as follows.
$g-\frac{k}{m}\sqrt{v}=0$
$\frac{k}{m}\sqrt{v}=g$
$\sqrt{v}=\frac{mg}{k}$
$v={\left(\begin{array}{c}\frac{mg}{k}\end{array}\right)}^{2}$
If $vo<{\left(\begin{array}{c}\frac{mg}{k}\end{array}\right)}^{2}$,then the object will fall faster and approaches the terminal velocity.
Similarly, if $vo>{\left(\begin{array}{c}\frac{mg}{k}\end{array}\right)}^{2}$, then the object will slow down to the terminal velocity.
Therefore, the terminal velocity is $v=\left(\begin{array}{c}\frac{mg}{k}\end{array}{\right)}^{2}$

Do you have a similar question?