problem02_48 solution

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

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2.48: a) From Eq. (2.8), solving for t gives (40.0 s m – 20.0 s m )/9.80 2 s m = 2.04 s. b) Again from Eq. (2.8), . s 12 . 6 s m 80 . 9 ) s m 0 . 20 ( s m 0 . 40 2 = - - c) The displacement will be zero when the ball has returned to its original vertical position, with velocity opposite to the original velocity. From Eq. (2.8),
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Unformatted text preview: . s 16 . 8 s m 80 . 9 ) s m 40 ( s m 40 2 =--(This ignores the t = 0 solution.) d) Again from Eq. (2.8), (40 s m )/(9.80 2 s m ) = 4.08 s. This is, of course, half the time found in part (c). e) 9.80 2 s m , down, in all cases. f)...
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This document was uploaded on 02/04/2008.

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