If the path length of a projectile launched from ground level is twice its range, what was the launch angle?

The projectile will have a parabolic arc, y(x) = a - bx^2, where I'm assuming the zenith happens to be at x=0. Because of this symmetry, the point of launch and the point at which it returns to the flat surface are just on opposite sides of 0. The arc length is then

integral_-c^c of sqrt(1 + (y')^2) dx

= integral_-c^c of sqrt(1 + (2bx)^2) dx

= c sqrt((2bc)^2 + 1) + arcsinh(2bc)/(2b)

The horizontal displacement is simply 2c. By assumption, we have

4c = c sqrt((2bc)^2 + 1) + arcsin(2bc)/(2b)

4 = sqrt((2bc)^2 + 1) + arcsinh(2bc)/(2bc)

The launch angle is just arctan(y'(-c)) = arctan(-2bc) = -arctan(2bc). Letting 2bc = x, the equation becomes

4 = sqrt(x^2 + 1) + arcsinh(x)/x

This has a solution at x ~= -3.2692023, at which point -arctan(x) ~= 1.27395 radians ~= 72.992 degrees.

If you let u = arctan(x), the equation above can be rewritten as

4 = sec(u) + cot(u) log(tan(u) + sec(u))

This is of course still horrific and essentially unsolvable algebraically.

The best angle for maximum distance is 45 degrees

However, unless you have missed out information. the RANGE of a projectile is the maximum distance it is able to travel so the projectile cannot have a path twice as long as the range