Shooting directly right:
Initially there is only velocity in the x direction (20 m/s). That x-direction velocity doesn't change as the cannon moves because there are no forces in the x direction acting on the cannonball. But there is a force acting on it in the y direction - gravity. The cannonball, in the y-direction, acts just like it would if it were dropped from the edge of the cliff. The kinematic equation yf - yi = vit + 1/2at2 can be used. The initial velocity in the y-direction is zero (the only velocity initially is in the x direction), and yf - yi = -14, and a is the acceleration due to gravity which is -9.81 m/s2 so -14 = 1/2(-9.81)t2 or t = √(14•2)/9.81 = √2.8542 = 1.689 s
Since the cannonball, in the x-direction, is always going 20 m/s, to find the distance it goes, multiply 20 m/s times the 1.689 s to get 33.8 m. If you only have 1 significant figure, you would need to round to 1 sig fig or distance = 30 m but I'm not sure if you've really got 1 sig fig or if you just typed the problem like that.