Chapter 5: Problem 14
In an amusement park ride, a rider suspended by cables swings back and forth from a tower. The maximum speed \(v\) (in meters per second) of the rider can be approximated by \(v=\sqrt{2 g h}\), where \(h\) is the height (in meters) at the top of each swing and \(g\) is the acceleration due to gravity \(\left(g \approx 9.8 \mathrm{~m} / \mathrm{sec}^2\right)\). Determine the height at the top of the swing of a rider whose maximum speed is 15 meters per second.
Short Answer
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Key Concepts
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