I figured there's got to be a way to calculate the altitude effectively. I thought it might have something to do with the descent, since the factors are more controlled than on the ascent (variable thrust hence variable velocity hence variable drag hence calculus). The answer was terminal velocity. Solve the drag equation for velocity, and call the drag mass x gravity since the rocket is in equilibrium. The result is the following equation:

v = [(2mg)/(pACd)]^(1/2) (I hate typing math. It always looks like someone just puked jibberish)

where v = velocity, m = mass of the rocket (0.550 kg), g = gravitational acceleration (9.8 m/s/s), p = the atmospheric density (1.2 kg/m^3 at the time), A = the area of the parachute (0.164 m^2), and Cd = the drag coefficient for our parachute (0.75 for a flat sheet). Plugging it all in yields v = 8.54 m/s.

I then went back to the video and timed the descent, which turned out to be 21 seconds. Using d = vt I was able to figure the rocket fell from 179 m, or 574 ft.

It is important to realize that this wasn't the maximum altitude the rocket reached, it was simply the point where the parachute popped out. The rocket may have fallen from a hundred or so feet beyond that.

If anyone finds a problem with any of my thinking, please let me know.

## 2/13/10

### Hey, I Know How High It Went!

by
DTH Rocket

Figure 1

Forces of drag and gravity are in equilibrium (Net force = 0)

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## 2 comments:

One minor problem I can think of, but it's small. Air density changes with altitude, so your rocket will be falling a little bit faster when it is closer to its apogee.

I thought of that, but when I looked into it the density change from 200 meters to 0 meters isn't significant. It would change the calculated altitude by only a couple feet. Thanks for the thought though.

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