# Vertical Motion in Air + Density [closed]

The fastest recorded skydive was by an Air Force officer who jumped from a helium balloon at an elevation of 103000 ft, three times higher than airliners fly. Because the density of air is so low at these altitudes, he reached a speed of 614 mph at an elevation of 90000 ft, then gradually slowed as the air became more dense. Assume that he fell in the spread-eagle position and that his low-altitude terminal speed is 125 mph. Use this information to determine the density of air at 90000 ft.

Thanks.

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## closed as too localized by David Z♦Oct 19 '11 at 17:59

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Hi John, and welcome to Physics Stack Exchange! This is not a homework help site, so we don't allow direct homework questions like this, especially without any work shown. But you can ask questions that arise in the course of doing a homework problem as long as they focus on the particular physical concept that is giving you trouble. There is more information on asking homework-related questions in this meta post. If you fix up this question (or your other one) to make it a conceptual one according to the guidelines, I would be happy to reopen it. – David Z Oct 19 '11 at 18:05

It's an interesting problem because the 614 mph obviously does not represent the terminal velocity at that elevation. You'll need to consider the net acceleration and a differential equation the utilizes the air density to get the drag force and is solvable for the position and velocity over time. Without loosing generalism I'll write this:

$$\frac{d^2 x(t)}{dt^2} = F_d \left(\frac{d x(t)}{dt},\rho (x(t))\right) - m g$$

If you find this unhelpful, that isn't particularly surprising. As formulated, this would be a rather difficult problem in my view. But I'll still write more about the drag term, $F_d$ which I've been writing as a function of a bunch of nasty stuff. A common form used for the drag force is proportional to the square of the velocity.

$$F_d \left(\frac{d x(t)}{dt},\rho (x(t))\right) = C \left( \rho(x(t)) \right) \left(\frac{d x(t)}{dt} \right)^2$$

I've left the drag coefficient in this form because in reality you could have some pretty sophisticated methods for finding it. The spread eagle position is relevant to its value, and density is as well. To a first approximation, I believe that $C$ will be directly proportional to $\rho$. Also, the form of $\rho(x)$, where $x$ is the vertical position, is complicated. Just look up information on the Earth atmosphere. You can make some progress assuming ideal gas laws. Wikipedia makes a good shot at this:

$$\rho(x) = \frac{p_0 \cdot \left(1 - \frac{L \cdot x}{T_0} \right)^\frac{g \cdot M}{R \cdot L} \cdot M}{R \cdot T}$$

Of course, there is much research left for you to do before applying this with confidence. You also will need to solve the differential equation I first wrote. You have some very straightforward boundary conditions to do that with.

$$x(0) = x_0$$ $$\left[ \frac{d x}{dt} \right]_{t=0} = 0$$

This is sufficient information to solve the differential equation (numerically at least) provided you've filled in all the functional forms I've left ambiguous. Then, the magnitude of the drag force function should be left unsolved for. This can be adjusted until you achieve the desired maximum speed. Unfortunately, that leaves the problem overdetermined since you also know the point of maximum speed as well as the drag coefficient from the sea-level terminal velocity. I don't really know how they want you to handle all those. You could do things like leave the coefficient on velocity in the drag equation unknown. Yes, that would blow the problem up to great complexity. In reality, since this is a homework problem, they probably want you to apply some very specific things from class. We don't answer homework problems exactly here anyway, I just wanted to share my initial thoughts and some governing equations. Good luck!

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