Postby **skeptical scientist** » Sun Nov 01, 2009 12:10 pm UTC

Well, without the max 5 second burn thing, here is what we get:

The total change in kinetic energy is equal to Mgh-W, where M is the mass of the rocket, and W is the work done by the rocket engines. The work W is equal to Mad, where a is the acceleration due to the rockets (20 m/s^{2}), and d is the total distance over which they are firing. Here Mgh=M*10,000 m^{2}/s^{2}, and the kinetic energy at landing can be at most M*12.5 m^{2}/s^{2} (the kinetic energy when speed is 5 m/s) so the work due to the rockets has to be at least M*9,987.5 m^{2}/s^{2}, so they must be exerted over a distance of 499.375 m. In order to minimize the time the rockets are firing (which serves to minimize both fuel use and total time, since the total time will be equal to the time the rockets are firing plus the time they are not, which must be within half of second of the time they are firing), we want to maximize the average speed over the time when the rockets are firing, so we want to fire the rockets at the last possible instant: when the ship is at height 499.375 m. So the optimal solution leaves the rockets off for 500.625 meters (10.006 seconds), and on for 499.375 meters (9.994 seconds), and uses slightly less than 10 units of fuel.

With the max 5 second burn thing, as you are interpreting it, we can get arbitrarily close to this by turning the rockets on for 4.997 seconds, then off for a nanosecond, and then on for another 4.997 seconds.

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson