the recoil velocity of the cannon is 0.12 m/s
Step-by-step explanation
The momentum of the projectile at the instant it was fired is equal to the product of its mass and its velocity (m*v). Then, by conservation of momentum, the cannon must move in the opposite direction with the same momentum
so
conservation of momemtum

replace
![\begin{gathered} m_(cannon)v_(cannon)=m_(cannonball)v_(cannonball) \\ (1,107.64\operatorname{kg})v_(cannon)=51.72\operatorname{kg}\cdot2.72(m)/(s) \\ 1107.64v_(cannon)=140.6784 \\ divide\text{ both sides by }1107.64 \\ (1107.64v_(cannon))/(1107.64)=(140.6784)/(1107.64) \\ v_(cannon)=0.12\text{ m/s} \end{gathered}]()
therefore,
the recoil velocity of the cannon is 0.12 m/s
I hope this helps you