[Fiddlerdave]

$this->bbcode_second_pass_quote('EnergyUnlimited', '')$this->bbcode_second_pass_quote('Beery1', 'W')ell done, kid!

And he's in Germany, where they tend to do schooling right, and not in the US or UK, where his genius might have been stifled by incompetent teachers or school officials.

[Timo]

$this->bbcode_second_pass_quote('ohanian', 'T')he problem he solved is as follows:
Let (x(t),y(t)) be the position of a particle at time t. Let g be the acceleration due to gravity and c the constant of friction. Solve the differential equation:
(x''(t)2 + (y''(t)+g)2 )1/2 = c*(x'(t)2 + y'(t)2 )
subject to the constraint that (x''(t),y''(t)+g) is always opposite in direction to (x'(t),y'(t)).
Finding the general solution to this differential equation will find the general solution for the path of a particle which has drag proportional to the square of the velocity (and opposite in direction). Here's an explanation how this differential equation encodes the motion of such a particle:
The square of the velocity is:
x'(t)2 + y'(t)2
The total acceleraton is:
( x''(t)2 + y''(t)2 )1/2
The acceleration due to gravity is g in the negative y direction.
Thus the drag (acceleration due only to friction) is:
( x''(t)2 + (y''(t)+g)2 )1/2
Thus path of such a particle satisfies the differential equation:
( x''(t)2 + (y''(t)+g)2 )1/2 = c*(x'(t)2 + y'(t)2 )
Of course, we also require the direction of the drag (x''(t),y''(t)+g) to be opposite to the direction of the velocity (x'(t),y'(t)). Once we find the intial position and velocity of the particle, uniqueness theorems tell us its path is uniquely determined.

[cipi604]

his father said that the mathematics of his son was beyond his knowledge so... this is basic mathematics in university, at age 16 yes it is far for his age but at the global level he has to learn years to come.

The calculations to get from A to B take a while ... with pen and paper, but everyone should understand it.

[Keith_McClary]

pelli makes it look easy:
$this->bbcode_second_pass_quote('', 'C')onsider a projectile moving in gravity with quadratic air resistance. The governing equations are
u' = -a * u * sqrt( u2 + v2 )
v' = -a * v * sqrt( u2 + v2 ) - g
where a is the coefficient of air resistance defined by |F| = ma|v|2 .
Cross-multiply and rearrange to find
a * sqrt( u2 + v2) * (uv'-vu') = gu'
Substitute v = su and separate variables:
a * sqrt( 1 + s2 ) * s' = g*u'/u3
Integrate both sides to get the answer:
g/u2 + a(v * sqrt( u2 + v2 )/u2 + arcsinh|v/u|) = const

[Outcast_Searcher]

$this->bbcode_second_pass_quote('dinopello', ' ')I think there are a lot of factors that will still require the numerical approach for real-world applications.

[vision-master]

Why is 'Three' so sacred?