I don't know what you are doing, either. It works the same way whether you use kinematics or energy.fishman100 wrote:1) Final velocity = sqrt(2*height*gravitational acceleration)
but in another post earlier in the year final velocity was said to have been the sqrt(height * (gravitational acceleration)).
Taken from: Link and Link
Don't know where the 2 in the first equation came from.
2) Time it takes for a falling mass to fall 1 meter = sqrt((2*distance)/9.8)), so at a 1m distance it takes about 0.452 seconds to drop
3) Velocity = same formula as final velocity; v = 4.427 m/s^2 --> 2.10 m/s
4) One of Balsa Man's runs at regionals: 2.61 seconds at 6.5 m --> 2.49 m/s
If you compare #3 and 4 they don't match up, and what puzzles me is that #3 theoretically the best possible time...
5) To calculate the force of gravity (gravitational acceleration) multiply 9.8 by the mass of the object. So for last year, that would be 9.8*2.5 = 24.5 N
Plug that into #1 and you get 7 m/s^2 --> 2.64 m/s, but that doesn't match with #3 or 4. Close but it's also theoretical, however it differs from the other theoretical equation (#3) by 0.54, which is a long time for something that should have the same results...
1: You calculated the time correctly, but the velocity incorrectly. Vf=at, so with t=0.452 Vf would be 4.43 m/s
2: mgh=1/2mv^2, so v^2=2gh and, given h=1, v=(2g)^1/2, which is 4.43 m/s.
Balsa Man, when you said the sliding mass produced faster times, you were comparing it with a fixed-in-the-center mass, right? It should not improve your time over a low fixed-in-the-rear mass, so the benefit you are getting over the latter is really in accuracy rather than speed, is it not?