Water jet on vane moving at 16 m/s

Assumptions: the vane moves at $V=16$ m/s in the direction of the jet; the stream leaves at C at $60^\circ$ above the horizontal relative to the vane; friction is neglected, so the relative speed is unchanged along the vane.

Relative velocity and flow rate

$$v_{rel}=v-V=48-16=32\ \text{m/s}$$

The mass of water striking a single moving vane per second is

$$\dot m=\rho A(v-V)=1000(4000\times10^{-6})(32)=128\ \text{kg/s}$$

Force on the stream

Apply the momentum equation in the frame moving with the vane (constant velocity, so Newton's laws hold):

$$\vec F=\dot m(\vec v_{C,rel}-\vec v_{B,rel})$$ $$\begin{aligned} F_x&=128\left(32\cos60^\circ-32\right)=-2048\ \text{N}\\ F_y&=128\left(32\sin60^\circ\right)=3547\ \text{N}\\ F&=\sqrt{2048^{2}+3547^{2}}=4096\ \text{N} \end{aligned}$$

This is the force the vane exerts on the stream. The stream exerts an equal and opposite force on the vane: $2048$ N forward (along the jet) and $3547$ N downward.

Power developed

Only the force component in the direction of the vane's motion does work:

$$P=F_xV=2048(16)=32\,768\ \text{W}\approx32.8\ \text{kW}$$

For a single moving vane, $P\propto(v-V)^2V$, and $dP/dV=0$ gives $V=v/3=16$ m/s. The given vane speed is exactly this value, so this is the maximum power.

Answer: resultant force on the stream $=4096$ N (components $-2048$ N and $3547$ N); maximum power $\approx32.8$ kW.