We rewrite our differential equation:

`(dv)/(dt)=k/mv^2-g`

in a more convenient form :

`(dv)/(dt)=g(k/(mg)v^2-1)`

Also for convenience, let *c* be defined as:

`c=sqrt((mg)/k`

So:

`1/c^2=k/(mg)`

We can rewrite our differential equation as:

`(dv)/(dt)=g(v^2/c^2-1)`

This is the same as:

`(dv)/(dt)=g((v^2-c^2)/c^2)`

Separating the variables:

`(c^2dv)/(v^2-c^2)=g\ dt`

Integrating both sides:

`c^2int(dv)/(v^2-c^2)=intg dt`

We multiply the fraction by `-1`, thus reversing the order of `(v^2-c^2)`, and also at the front to compensate, so the substitution step coming up later is possible. (Otherwise, we would be trying to find the log of a negative number when finding `K`.)

`-c^2int(dv)/(c^2-v^2)=int g dt`

To perform this integration, we could either:

- Factor the denominator, use partial fractions and then integrate (it needs the logarithm form), or
- Use a table of integrals (integral #13), (easier); or
- Use a computer algebra system, like Scientific Notebook. (easiest)

We integrate and obtain:

`(-c^2)1/(2c)ln((c+v)/(c-v))= g t +K`

"*K*" is the constant of integration.

At `t = 0`, `v = 0` and after substituting, we conclude `K = 0`. Therefore:

`-c/2ln((c+v)/(c-v))=g t`

Now we solve for *v*.

Multiply both sides by `-2` and divide by *c*:

`ln((c+v)/(c-v))=-(2g t)/c`

Take *e* to both sides to remove the logarithm:

`(c+v)/(c-v)=e^(-2g t"/"c)`

Multiply out and solve for *v*:

`c+v=(c-v)e^(-2g t"/"c)`

`c+v=ce^(-2g t"/"c)-ve^(-2g t"/"c)`

`ve^(-2g t"/"c)+v=ce^(-2g t"/"c)-c`

`v=c(e^(-2g t"/"c)-1)/(e^(-2g t"/"c)+1)`

This is the velocity of our object at time *t*.

First, we find *c* for our given situation. This was the expression for *c*:

`c=sqrt((mg)/k)`

We use the given mass and the coefficient of drag for the skydiver.

mass =

m= 80 kgcoefficient of drag =

k= 0.2

So

`c=sqrt((mg)/k)`

`=sqrt((80xx9.8)/0.2)`

`=62.6`

Next, we substitute the given time value (*t* = 5) and *c* = 62.6 to find the required velocity:

`v=c(e^(-2g t"/"c)-1)/(e^(-2g t"/"c)+1)`

`=62.6(e^(-2(9.8)(5)"/"62.6)-1)/(e^(-2(9.8)(5)"/"62.6)+1)`

`=-40.9580`

The units are ms^{-1}, so the velocity at time *t* = 5 s is approximately 147 km/h (1 ms^{-1} = 3.6 km/h), in the downward direction.

The expression for velocity at time *t* we found earlier:

`v=c(e^(-2g t"/"c)-1)/(e^(-2g t"/"c)+1)`

As *t* → ∞, the value of the fraction approaches −1, since *e*^{-2gt/c} → 0, giving us the terminal velocity `v=-c`.

So

`c=sqrt((mg)/k)`

is the **terminal velocity** for the falling object (in the downward direction).

**Note: **We could have obtained the above expression without knowing the expression for velocity at time *t*, by simply noting the velocity of the object reaches terminal velocity when the acceleration is 0.

That is, solving the following acceleration expression to find the velocity:

`a=k/mv^2-g=0`

This gives terminal velocity

`v=sqrt((mg)/k)`

We already found the expression for *c* (which is the terminal velocity) in Part (b)

`c=sqrt((mg)/k)`

`=sqrt((80xx9.8)/0.2)`

`=62.6`

So the terminal velocity is approximately 225 km/h, since 1 ms^{-1} = 3.6 km/h.

The graph of the velocity against time shows that it takes around 15 seconds to reach (or "get very close to") the terminal velocity:

Velocity graph `v=225(e^(-2(9.8) t"/"62.6)-1)/(e^(-2(9.8) t"/"62.6)+1)`, showing the point `(5, -147)`.

**Note:**

- The graph shows that the terminal velocity is never actually reached - the skydiver's velocity just gets closer and closer to that velocity.
- The graph includes the point representing the velocity at time
*t*= 5, found earlier. - Actually, the air resistance increases as the air gets more dense nearer the Earth's surface. We have assumed it remains constant for this problem.
- The human sky diver can change
*k*easily by either spreading their arms and legs (which will slow them down), or diving down with arms and legs tightly together (which will increase their speed)

Easy to understand math videos:

MathTutorDVD.com