1. The problem statement, all variables and given/known data A cannon with a muzzle speed of 992 m/s is used to start an avalanche on a mountain slope. The target is 1900 m from the cannon horizontally and 793 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.) 2. Relevant equations Xf= xi+vit Yf=yi+vit+1/2at2 vfy^2=viy^2+2a(yf-yi) x=cos(theta)992m/s Y=sin(theta)992m/s 3. The attempt at a solution ok this one i can get all the equations setup. xf= cos(theta)992m/s 793=(sin(theta)992m/s)t + 1/2(-9.80)t2 using the x eqaution i can solve for t. t=1900/cos(theata)992m/s but once i place it into the eqaution, it gets very messy and i get stuck on the arithmatic. 793= (sin(theta)992m/s)(1900/cos(theata)992m/s)+1/2(-9.80)(1900/cos(theata)992)^2 this is where i am stuck.