We got to use the resources of the tech-ed department to paint the rocket. We only spent about three hours total on it, which wasn't enough priming and sanding to remove the spiral from the phenolic airframe, but the end result was still pretty good. We used black and gold, the school colors. Its black with a double gold spiral from nose to aft. Silly me, though. I seem to have left my camera at school! Come back on Monday and I'll have some great pictures.

Here's the solution to #4 on my pop quiz:

Sound intensity I = P/(4*pi*r^2), P is power and r is the sound distance from the source. In this case, r = d.

Apparently the actual sound intensity didn't matter because they gave us I and 1/3I. So I plugged some random number like 1 in for I. I_{1} = 1 W/m^2 and I_{2} = .333 W/m^2. Rearanging the intensity equation for power is P = 4*pi*d^2*I. d_{1} = 562 meters. P = 4*pi*562^2*1 = 3.97x10^6 W. I then solved the intensity equation for d so I could find d_{2}. d = (P/(4*pi*I))^(1/2). Plugging 3.97x10^6 in for P and .333 in for I you get 973.4 meters. Once you have this, you can use the kinematic equation d = 1/2at^2 to find t_{1} and t_{2}. Solving for t you get t = (2d/a)^(1/2). t_{1} = 5.79 seconds and t_{2} = 4.40 seconds. The change in time from the first sound emmission to the second is t = t_{1} - t_{2} = 1.39 seconds. But this isn't the final answer, because the question asks for the time between each of the measurements below at the ground-based monitoring station. Sound takes time to travel through the air, so you have to find the amount of time each sound wave took to go from the altitude they were emitted at and the station. t = d/v, so t_{3} = 562m/343m/s = 1.64 seconds, t_{4} = 973m/343m/s = 2.84 seconds. So to find the time between the two measurements you just have to sit there and think about it for a while, and finally you come up with t = 2.84 sec + 1.39 sec - 1.64 sec = **2.59 seconds**

I thought that was a fun problem.

## 2/27/10

### A Slick Paint Job

by
DTH Rocket

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