#### a. General Formula for Velocity of an Object Falling in Air

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`

To perform this integration, we could either:

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

We obtain:

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

"K" is the constant of integration.

At `t = 0`, `v = 0` so `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:

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

This is the velocity of our object at time t.

#### b. Velocity for Particular Sky Diver at Specific Time

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 kg

coefficient 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.

#### c. Terminal velocity

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)`

#### d. Terminal Velocity for Skydiver Example

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 (1 ms-1 = 3.6 km/h).

The graph of the velocity against time shows that it takes around 15 seconds to reach the terminal velocity:

Note:

1. The graph shows that the terminal velocity is never actually reached - the skydiver's velocity just gets closer and closer to that velocity.
2. The graph includes the point representing the velocity at time t = 5, found earlier.
3. Actually, the air resistance changes as the air gets more dense nearer the Earth's surface. We have assumed it remains constant for this problem.
4. 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)