I can see where the formula comes from (and I'm assuming that the initial velocity was 5, as that seemed to work to get the form of equation you've shown us). I come up with an equation identical to yours, except the 39.2 is replaced by 2g(h-y) where g is acc due to grav, h is initial launch height, and y is height of the object at some time...
x = (5 cos{@}) (5 sin{@} +
√{25 (sin{@})^2 + 2g(h-y)}) / 9.8
This means that for yours to work, the y value is precisely 2 units less than the original height. This is perfectly reasonable. If we want to turn this 'x' into range of the projectile, R, we must look for the case when y=0, so this becomes:
R = (5 cos{@}) (5 sin{@} +
√{25 (sin{@})^2 + 2gh}) / 9.8
Now we have this, we can see when we graph it as x(@) vs @ that increasing 'h' merely increases the amplitude of the function and moves it slightly towards the left. If we look at the maximum points of the graph, (best to use some sort of graphing/math s/w, I used mathematica) by differentiating wrt @, different values of h show different max values. I tested out h=0,10,20,30,40,50,100; and each time the max point moved to the left a bit.
As far as I can tell, as the height increases, the range increases, regardless of the angle; however, as the height increases, the angle for the maximum range seems to tend towards 0, which is rediculous...
But here's a graph of its derivative with h=0 (right-most),10, 50, 100, 200 (left-most). This clearly shows that the extreme point approaches 0 as R increases.
Hope this helps... tho it may not be right coz you wouldn't expect that if you could throw the ball an infinite distance straight up, it would go an infinite distance in any other direction in the horizontal plane...
Any thoughts, fellow numeromaniacs?
