First understand the only work done on the projectile is done by the force of gravity F acting over some distance. There is no work horizontally as there are no horizontal forces if we discount air drag. As F is strictly up and down, the distances it works over are also just up and down... vertical.
So we start with H = h + Uy^/2g as the max height above ground, where h = 6.6 m and Uy = 6.6 mps. If we discount the work mgh to raise the projectile to the muzzle of the cannon, then work to reach H is mgH - mgh = 1/2 mUy^2 which is just the initial kinetic energy from vertical speed Uy at the muzzle.
The work in descent will be mgh + 1/2 mUy^2 from the conservation of energy. In this case we include the work mgh to get the projectile up to height h. Thus the total work is W = mgh + 2(1/2 mUy^2) = m(gh + Uy^2) = .22*(9.81*6.6 + 6.6^2) = 23.83 Joules. ANS.
V = sqrt(Vy^2 + Ux^2) the impact velocity; where Ux = 6.2 mps, and Vy^2 = 2gH = 2g(h + Uy^2/g) = 2*9.81*(6.6 + (6.6)^2/9.81) = 216.6 In which case V = sqrt(216.6 + 6.2^2) = 15.97 mps at omega = arctan(Vy/Ux) = ATAN(sqrt(216.6)/6.2) = 67.2 deg re the horizontal. 15.97 mps @ 67.2 deg ANS.
Time to reach H is tu = Uy/g = 6.6/9.81 = 0.67 seconds; time to fall from H is td = sqrt(2H/g) = sqrt(2*(6.6 + 6.6^2/9.81)/9.81) = 1.5 seconds. Total flight time T = tu + td = .67 + 1.5 = 2.17 seconds, so the range R = UxT = 6.2*2.17 = 13.45 meters ANS.
This is a good test of your understanding of the physics. First, work is force over distance; so there must be a force and there must be a distance. The only force is the vertical force of gravity, discounting drag; so the only work done is: 1. to slow the vertical speed down to zero at H and 2 to speed the vertical speed up to Vy when falling from H.
Second, the range R is based on the time of flight T and the horizontal speed Ux = Vy throughout the flight. So we need to add the time to ascend to H and the time to descend from H to get the total flight time T. Because the projectile in this case started from h > 0, the tu < td where td is the time to fall from H > h. So it's pretty much a given that you will need to find H = h + Uy^2/g to find td = sqrt(2H/g), the time of fall from H.