Maximum horizontal displacement, again.

[mardlamock]

Hello, I ve been doing a bit more on those functions that express the maximum horizontal distance a rocket could travel without considering the sphereness of the earth and the atomsphere. Basically instead of looking for the rocket to stay at a same height through the time the fuel lasts I want to accelerate in a 45 degree angle as much as I can. In order to do so the rocket needs shed a bit of mass if the vertical component of its force at a 45 degree angle is smaller than the mass of the rocket times 9.81. So I ended up with this: http://latex.codecogs.com/gif.latex?%5CTheta%20%28t%29%3D%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20sin%5E-1%28%28m-r*t%29*g/f%29%20%26%20if%5C%2C%20t%3C%28-f/sqrt%282%29+g*m%29/%28g*r%29%5C%5C45%5E%7B%5Ccirc%20%7D%20%26%20if%5C%2C%20t%3E%28-f/sqrt%282%29+g*m%29/%28g*r%29%20%5Cend%7Bmatrix%7D%5Cright

Now, the horizontal acceleration as a function of t is: http://latex.codecogs.com/gif.latex?a_%7Bh%7D%28t%29%3D%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%5Cfrac%7Bcos%28sin%5E-1%28%28m-r*t%29*g/f%29%29%29*f%7D%7B%28m-r*t%29%7D%20%26%20if%20%5C%2C%20t%3C%28-f/sqrt%282%29+g*m%29/%28g*r%29%5C%5C%20%5Cfrac%7B1/sqrt%282%29*f%7D%7B%28m-r*t%29%7D%26%20if%20%5C%2C%20t%3E%28-f/sqrt%282%29+g*m%29/%28g*r%29%20%5Cend%7Bmatrix%7D

Where (-f/sqrt(2)+g*m)/(g*r) is equal to the time it takes for the rocket s force at 45 degrees to equal the weight of the rocket.

Now, v(t) would be looking something like this right?:http://latex.codecogs.com/gif.latex?v_%7Bh%7D%28t%29%3D%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%5Cint_%7B0%7D%5E%7Bt1%7D%20%5Cfrac%7Bcos%28sin%5E-1%28%28m-r*t%29*g/f%29%29%29*f%7D%7B%28m-r*t%29%7D%20%26%20if%20%5C%2C%20t%5Cleq%20%28-f/sqrt%282%29+g*m%29/%28g*r%29%20%5C%5C%20%5Cint_%7B0%7D%5E%7B%28-f/sqrt%282%29+%28g*m%29/%28g*r%29%20%7D%20%5Cfrac%7Bcos%28sin%5E-1%28%28m-r*t%29*g/f%29%29%29*f%7D%7B%28m-r*t%29%7D+%5Cint_%7B%28-f/sqrt%282%29+g*m%29/%28g*r%29%20%7D%5E%7Bt2%7D%5Cfrac%7B1/sqrt%282%29*f%7D%7Bm-r*t%7D%26%20if%20%5C%2C%20t%20%3E%20%28-f/sqrt%282%29+g*m%29/%28g*r%29%20%5Cend%7Bmatrix%7D%5Cright.

Now, wolfram can integrate that and actually has, the problem is that it is a ridiculously long function and im not even describing the parabolic motion it follows once the fuel has been depleted. Would there be a simpler way to do it?

If there is anything I wrote in a weird way please tell me so I can explain it. Thanks!

[mardlamock]

No one gonna answer?

[K^2]

In general, it's probably even worse, because I'm not sure that 45° angle is your best bet. But I'm pretty sure that full throttle and constant angle of the rocket is. So the problem reduces to a) Finding where the rocket ends up when the engine cuts off, and Computing the range from that using range formula, and optimizing result by varying the angle. Which seems to be the way you were going about it.

Final velocity is pretty easy. You just apply rocket formula, and subtract off the amount of impulse you are going to lose to gravity. That might make for a simple v(t) than what you have right off the bat. What's a bit more complicated is getting final height. You can, of course, integrate over velocity to get final position, but that doesn't sound fun. But you can make it much simpler. Again, split velocity into the pure rocket formula component and gravity component. After all, a gravity problem is equivalent to an accelerated frame problem. Remember the monkey and the hunter problem? Anyways, just integrate rocket formula separately, add in the free-fall, and you should get final position as a much neater formula.