problem03_44

University Physics with Modern Physics with Mastering Physics (11th Edition)

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3.44: a) Using generalizations of Equations 2.17 and 2.18, 2 2 0 3 3 0 , t t v v t v v y y x x γ α β - + = + = , and 3 6 2 2 0 4 12 0 , t t t v y t t v x y x γ β α - + = + = . b) Setting 0 = y v yields a quadratic in 2 2 0 0 , t t v t y γ β - + = , which has as the positive solution [ ] s, 59 . 13 2 1 0 2 = + + = γ β β γ v t keeping an extra place in the intermediate calculation. Using this time in the expression for y ( t ) gives a maximum height of 341 m.
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Unformatted text preview: y = involves solving another quadratic, 2 6 2 t t v y γ β-+ = (note that the root = t has been factored out). Solving for t gives s 73 . 20 = t (keeping the extra figure in the intermediate calculation), at which time km 5 . 38 = x ....
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