At what angle, above the horizontal, should the cannon be fired?

Question

A cannon with a muzzle speed of 995 m/s is used to start an avalanche on a mountain slope. The target is 2000 m from the cannon horizontally and 809 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.)
what degree?

Answer 1

There are two possible angles: 22.593 deg and 89.431 deg.
theta = arctan[(v0^2 +/- sqrt{v0^4 - g*(g*range^2+2*yt*v^2)}) / (g*range)] =
89.4305169604134 22.5928151618953 deg
Equation derived from:
yt = range*tan(theta) - g*range^2*(1+tan(theta)^2)/(2*v0^2)
where yt is target height, from the ref. web page.
Assumed g = 9.8 m/s^2.
Note: Yahoo currently has a bug that prevents posting the URL unless a space is inserted. (You get a "999" error.) To reach the ref., paste it into your browser's address bar and remove the space before "org".