Answer
The total energy stays the same, as discussed on page 118.
In other words, at the beginning, the banana's mechanical energy is all in the form of PE = $mgh$, where h = 0 at the surface of the river.
This energy is transformed into KE as the banana falls. When the banana is about to hit the water, h = 0 and the energy is all in the form of KE, $\frac{1}{2}mv^{2}$.
This expresses conservation of energy.
$$mgh=\frac{1}{2}mv^{2}$$
Solve for the speed just before hitting the water.
$$2mgh=mv^{2}$$
$$2gh=v^{2}$$
$$\sqrt{2gh}=v$$
We have proven the desired result.
