Consult ConceptualExample 9 in preparation for this problem. InteractiveLearning

Question

Consult ConceptualExample 9 in preparation for this problem. InteractiveLearningWare 6.3 also provides useful background. The drawingshows a person who, starting from rest at the top of a cliff,swings down at the end of a rope, releases it, and falls into thewater below. There are two paths by which the person can enter thewater. Suppose he enters the water at a speed of 17.4 m/s via path1. How fast is he moving on path 2 when he releases the rope at aheight of 2.31 m above the water? Ignore the effects of airresistance.

Explanation / Answer

total energy = PE + KE if he enters water with 17.4 m/s (via path 1), total energy = 0 + (1/2)m(17.4)^2 via path 2, if his speed is v at a height h = 2.31m abovewater, total energy = PE + KE = mgh + (1/2)mv^2 so, since energy is conserved, mgh + (1/2)mv^2 = (1/2)m(17.4)^2 v^2 = (17.4)^2 - 2gh = (17.4)^2 - 2(9.8)(2.31) v = 16.05 m/s

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