# Felix's Parachute Jump

##### Stage: 5 Challenge Level:
We had a number of really good attempts at solutions sent to us. Well done to Thomas, Christian, Nirali, Omar and Seemanta!

We also received a very nice solution from Margaret, which you can see here: .pdf

You may notice that using SUVAT equations to answer the second and third question gives you some slightly dubious answers! Mousam from New College (Swindon) used a more sophisticated approach:

First, we need to assume few things. Due to these assumptions, the terminal velocity of Felix is unlikely to be greater than the speed of sound. However, we should still get an approximate answer.

We are going to start off with an assumption. The acceleration of Felix due to the gravity of earth is constant. This acceleration is going to be the mean of the acceleration at sea level and the acceleration at 39,000m. The acceleration at 39,000m can be worked out by using the following:

$F = ma$   and   $F = \frac{GMm}{r^2}$

where $F$ is the force due to gravity, $M$ is the mass of the Earth, $m$ is the mass of Felix, $G$ is the Universal Gravitational constant and $r$ is the distance between the centre of the mass of the Earth and the centre of the mass of Felix.

$$ma = \frac{GMm}{r^2}$$
$$\Rightarrow a = \frac{GM}{r^2} = \frac{(6.67\times10^{-11})(6.0\times 10^{24})}{(3.9\times10^4 + 6.4\times 10^6)^2}$$

(Since $r$ is the distance between the centre of the mass of the Earth and the centre of the mass of Felix, we need to add the radius of the Earth and the distance between the surface of the Earth and Felix to get the separation.)

So at 39,000m, $a = 9.65\, \text{ms}^{-2}$ (approx.)

Therefore, average acceleration $\frac{9.65 + 9.81}{2} = 9.73\, \text{ms}^{-2}$.

We are going to take this value, $9.73\, \text{ms}^{-2}$, and treat it as a constant acceleration to make our mathematics easier.

Next, we need to know how to find the terminal velocity of an object. This
is given by the following equation:

$$V^2 = \frac{2mg}{\rho AC_d}$$
where $V$ is the terminal velocity of the object, $m$ is the mass of the object, $g$ is the acceleration due to gravity, $\rho$ is the density of the fluid through which the object is travelling through, $A$ is the projected area of front of the object and $C_d$ is the drag coefficient of the object.

Some of these can only be obtained through research such as the mass of Felix and the drag coefficient of Felix. Others, like the density of the air through which the Felix is travelling through must be worked out separately. If we were to make a list of things we need to research and things we need to calculate, it would look something like this:

Things we need to research : Drag coefficient of Felix, mass of Felix.
Things we need to calculate : Fluid density, projected area of the front of Felix, terminal velocity.

Drag Coefficient :
Drag coefficient of a normal person is about 1-1.3. Since Felix is carrying all that gear and chest packs, let's assume that his drag coefficient is at the higher end (1.3).

Mass of Felix :
On the official website of Red Bull, they have stated that including equipment, Felix had a mass of 260 pounds which is about 120kg.

Projected area of the front of Felix:
When upright, a normal body takes around $0.1\, \text{m}^2$ but since he was tumbling a little bit, we can assume the projected area of the front of Felix to be about $0.15\, \text{m}^2$.

Fluid Density:
To calculate this, we need a bit more mathematics. Fluid density can be worked out by using the following equation:

$$\rho = \frac{p}{RT}$$
where $\rho$ is the fluid density, $p$ is the absolute pressure, $R$ is the specific gas constant and $T$ is the absolute temperature.

Further research on the internet reveals that the absolute pressure at 39,000m is only about 0.33% of the absolute pressure at sea level (101000 pa). Given this information, We can work out the absolute pressure at 39,000m.

$p = \frac{0.33\times 101000)}{100} = 333.3\, \text{Pa}$ (approx.)

The specific gas constant is $8.31\, \text{JK}^{-1}$ and further research tells us that at 39,000m, the temperature was -14 degrees Fahrenheit. However, we need the absolute temperature therefore the absolute temperature at 39,000m is 247.59 Kelvin.

We now have everything we need to work out $\rho$.

$\rho = \frac{p}{RT} = {333.3}{8.31\times 247.59} = 0.162\, \text{kgm}^{-3}$ (approx.)

With this last piece of data, we can now finally work out the terminal velocity.

$V^2 = \frac{2mg}{\rho AC_d} = \frac{2\times 120 \times 9.73}{0.162\times 0.15\times 1.3} = 7.4\times 10^4\, \text{m}^2 \text{s}^{-4}$ (approx.)

$\Rightarrow \, V = 270\, ms^{-1}$

This means that the terminal velocity of Felix, given that our assumptions are correct, is about $270\, \text{ms}^{-1}$ which is below the speed of sound at 39,000m (which is approximately $316\, \text{ms}^{-1}$).

Since we now know the terminal velocity, we can use the equations of motion with constant acceleration to find the time it would take Felix to reach the surface of the Earth if he had not opened his parachute. First of all, let's find out the time taken for Felix to reach his terminal velocity of $270\, \text{ms}^{-1}$ given that the acceleration is $9.73\, \text{ms}^{-2}$.

$v = u + at$, where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration due to gravity and $t$ is the time taken.
$t = (v - u)a = \frac{270}{9.73} = 28\, \text{seconds}$ (approx.)

Now let's find out the distance travelled in these 28 seconds so that we can work out how high he is from the ground.

$s = ut + \frac{at^2}{2}$, where $s$ is the displacement of the object from its initial position.
So $s = 0 + \frac{9.73\times 28^2}{2} = 3800\, \text{m}$ (approx.)

Since he has travelled 3800m in 28 seconds, we can work out the distance $d$ left between Felix and the ground: $d = 39000 - 3800 = 35200\, \text{m}$.

Now that Felix has reached his terminal velocity, we are going to assume that his velocity does not change for the rest of the fall. This means that we will ignore the increase in air resistance and the fact that air density increases with decrease in height. This will make the mathematics a bit easier because if we took into account the increase in air resistance, we would have to use differential equations.

Since he is travelling at $270\, \text{ms}^{-1}$ and that the distance between Felix and the ground is 35200m, we can do a simple calculation to work out the time taken for Felix to reach the ground.
$v = \frac{s}{t}$, where $v$ is the velocity of an object, $s$ is the displacement of the object and $t$ is the time taken.
$t = \frac{s}{v} = \frac{35200}{270} = 130\, \text{s}$.

We can now add the time taken by Felix to reach his terminal velocity and the time taken to reach the ground after Felix reached his terminal velocity to get the total time of free fall.

Total time = 28 + 130 = 158s (approx.)

In conclusion, it would take Felix about 2 minutes and 38 seconds to reach the surface of the Earth if he fell from 39,000m without opening his parachute.

Well done Margaret and Mousam!