Albert's skydive

Here's one way to solve this problem. We start with Bernard's statement:

(1) $\quad v(t+dt)\quad -\quad v(t)\quad =\quad gdt\quad -\quad \mu v(t)dt$

We can rearrange (1) by dividing both sides by the increment 'dt'

(2) $\quad \frac { v(t+dt)-v(t) }{ dt } \quad =\quad g\quad -\quad \mu v(t)$

taking the limit as 'dt' goes to 0, (2) yields (3) which is a first-order Ordinary Differential Equation:

(3) $\quad \frac { dv }{ dt } \quad =\quad g\quad -\quad \mu v(t)$

We want to solve for $v(t)$ : First we look for a general solution (i.e a solution to the homogenous equation ${ E }_{ H }$ )

( ${ E }_{ H }) \quad \frac { dv }{ dt } \quad +\quad \mu v(t)\quad =\quad 0$

We're looking for a function which first derivative is equal the same function times another function (here $-\mu$ is a constant function).

Since \frac { d }{ dt } { e }^{ f(t) }\quad =\quad f'(t){ e }^{ f(t) } exponentials are known to have this property.

Therefore { v }_{ H }(t)\quad =\quad A{ e }^{ M(t) },\quad with\quad M(t)\quad =\quad \int _{ 0 }^{ t }{ -\mu dt }

So the general solution to ( ${ E }_{ H }$ ) is

${ v }_{ H }(t)\quad =\quad A{ e }^{ -\mu t }$ , where 'A' is a constant defining the initial conditions of the system.

Now all we have to do is to find a particular solution to (3) , you should always look for a constant solution first. Here a constant solution exists:

${ v }_{ p }(t)\quad =\quad \frac { g }{ \mu }$

The complete solution is the sum of the general and particular solutions:

$v(t)\quad =\quad \frac { g }{ \mu } +\quad A{ e }^{ -\mu t }$

Now requiring that $v(0)={ v }_{ 0 }=0$ implies $A=\frac { -g }{ \mu }$

So $v(t)=\frac { g }{ \mu } (1-{ e }^{ -\mu t })$

The terminal speed, ${ v }_{ M }$ , is given by

${ v }_{ M }\quad =\quad \underset { t\rightarrow +\infty }{ lim } v(t)\quad =\quad \frac { g }{ \mu }$

And so we obtain the ratio between the speed and the terminal speed:

$\frac { v }{ { v }_{ M } } (t)\quad =\quad 1-{ e }^{ -\mu t }$

Which when evaluated at t=10s gives:

$\frac { v }{ { v }_{ M } } \quad$ = $\boxed{0.83}$