# How high can water rise up your arm hanging in the water from a moving boat? (Bernoulli's principle)

I have the following problem where you have to apply Bernoulli's principle, which asks:

How high can water rise up one's arm hanging in the river from a boat moving at $$u_1 = 1 \text{ ms}^{-1}$$?

(That is, you have your arm hanging over the side of the boat partially submerged in the water, and want to find how high the water would rise relative to the initial level of submersion due to the speed of the water flow).

So far I have tried considering the rest frame of the boat (so the water is moving at $$1 \text{ ms}^{-1}$$) and applying Bernoulli's principle to a streamline on the water's surface. Taking $$h_1$$ to be the original height of the water level and $$h_2$$ to be the height of the water when it reaches your arm, you get $$\frac{1}{2} \rho u_1^2 + p_1 + gh_1 = \frac{1}{2} \rho u_2^2 + p_2 + gh_2$$ and the pressure $$p_1$$ and $$p_2$$ should be the same at both points since we're at surface level, so we get $$h_2-h_1 = \frac{1}{2g} \rho(u_1^2-u_2^2).$$ We know $$u_1$$, but I'm not sure how you'd find $$u_2$$, especially considering how there isn't a pipe wall (or anything like that) above the water to constrain its volume, so applying conservation of mass etc. doesn't seem to work. Is this the right way of setting up the problem, and how do you find $$h_2$$?

You can consider $$u_2 \approx 0$$, so that gives you:
$$h_2-h_1 \approx \frac{1}{2g} \rho u_1^2$$
• Why would $u_2$ be very small? Commented Feb 6, 2022 at 18:21
• Imagine we replace the arm with a cylinder (easier to 'see' in your mind). As the boat proceeds at $u_1$ we insert the cylinder into the water. The water now 'creeps' up at the front side of the cyl. Soon this process is completed and the water no longer creeps up any further: its velocity is effectively zero, at $h_2$.