problem02_83 solution

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

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2.83: a) From Eq. (2.14), with v 0 =0, s. m 59 . 7 m) 640 . 0 )( s m 0 . 45 ( 2 ) ( 2 2 0 = = - = y y a v y y b) The height above the release point is also found from Eq. (2.14), with , and 0 s, m 59 . 7 0 g a v v y y y - = = = m 94 . 2 ) s m 80 . 9 ( 2 ) s m 59 . 7 ( 2 2 2 2 0 = = = g v h y (Note that this is also (64.0 cm) ( 29 g 2 s m 45 .The height above the ground is then 5.14 m. c) See Problems 2.46 & 2.48 or Example 2.8: The shot moves a total distance
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Unformatted text preview: 2.20 m –1.83 m = 0.37 m, and the time is s. 60 . 1 ) s m 80 . 9 ( m) 37 . ( ) s m 80 . 9 ( 2 ) s m 59 . 7 ( ) s m 59 . 7 ( 2 2 2 = + +...
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• Total Distance, release point

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