Conceptual Physics (12th Edition)

Published by Addison-Wesley
ISBN 10: 0321909100
ISBN 13: 978-0-32190-910-7

Chapter 7 - Think and Solve: 50

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.
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