## Conceptual Physics (12th Edition)

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.