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Here's a simple formula and a rapidly convergent
algorithm for your case. You can work it out easily for yourself if
you just remember that horizontal distance traveled is a fine proxy
for the elapsed time:
1 Define a parameter q=2gH/V^{2} (V^{2} is the TeX way to
write the exponent 2; V is the magnitude of the intitial velocity
vector)
2. The launch angle a for maximum range is given by:
Cos(2a)= qSin(a)/[Sin(a)+A)
where A^{2} = Sin^{2}(a) +q
3. For the algorithm, put a starting guess for a on the right hand
side.
The value you get for Cos(2a) gives you an improved value for a. 45
degrees is a good starting guess. This procedure converges to three
places in one iteration for q=.4. The angle for max range in that
case is 40.2 degrees.
There may be a condition on the starting angle for values
of q>3. The condition may be that Sin(a)<1/sqrt(q-2). I'll leave it
for others to explore this condition.
Regards,
Jack