problem03_56

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

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Unformatted text preview: 2 3.56: The equations of motions are y = (v0 sin )t - 1 / 2 gt and x = (v0 cos )t , assuming the match starts out at x = 0 and y = 0 . When the match goes in the wastebasket for the minimum velocity, y = 2 D and x = 6 D . When the match goes in the wastebasket for the maximum velocity, y = 2 D and x = 7 D . In both cases, sin = cos = 2 / 2. . To reach the minimum distance: 6 D = 22 v0t , and 2 D = 22 v0t - 1 gt 2 . Solving the first 2 equation for t gives t = 2D = 6D - 1 g 2 ( 6D 2 v0 ) . Solving this for v 2 6D 2 v0 . Substituting this into the second equation gives 0 gives v0 = 3 gD . 2 2 To reach the maximum distance: 7 D = equation for t gives t = 2D = 7 D - 1 g 2 v0t , and 2 D = 2 2 v 0 t - 1 gt 2 . Solving the first 2 ( 7D 2 v0 ) . Solving this for v 2 7D 2 v0 . Substituting this into the second equation gives 0 gives v0 = 49 gD / 5 = 3.13 gD , which, as expected, is larger than the previous result. ...
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