What distance can a cannonball traverse thru water without losing too much kinetic energy? For a back-of-the-envelope calculation we start from the observation that this distance scales with the ratio of the kinetic energy of the cannonball and the drag force exerted on the cannonball.
Let's denote the ball's radius by $R$, its speed by $v$, and its mass density by $\rho_{ball}$. The kinetic energy $E_k$ equals $\frac 1 2 M v^2 = \frac{2 \pi}{3} \rho_{ball} R^3 v^2$.
The drag force $F_d$ is given by $\frac 1 2 C_d \rho_{water} v^2 A = \frac {\pi}{2} C_d \rho_{water} v^2 R^2$. Here, $C_d$ denotes the drag coefficient for a sphere.
The maximum distance $L _{max}$ that can be traversed by a cannonball $L_{max} = E_k/F_d$ is therefore $\frac 4 3 \frac {R}{C_d} \frac {\rho_{ball}}{\rho_{water}}$. For typical values ( $\frac{\rho_{ball}}{\rho_{water}} < 8$ and $C_d > 0.1$, see here), we find $L_{max} < 100 R$.
In other words, a cannonball loses much of its kinetic energy when it traverses a layer of water larger than about fifty times its diameter.
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