problem05_p106

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

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5.106: (a) To find find the maximum height and time to the top without fluid resistance : s 61 . 0 s m 8 . 9 s m 0 . 6 0 m 1.8 or m 84 . 1 ) s m 8 . 9 2( s) m 0 . 6 ( 0 2 ) ( 2 2 0 2 2 2 0 2 0 0 2 0 2 = - - = - = = - - = - = - - + = a v v t a v v y y y y a v v y y y y (b) Starting from Newton’s Second Law for this situation kv mg dt dv m - = we rearrange and integrate, taking downward as positive as in the text and noting that the velocity at the top of the rock’s “flight” is zero : 386 . 1 ) 25 . 0 ln( s m 0 . 2 s m 0 . 6 s m 0 . 2 ln ln ) ln( t t 0 t 0 t - = = - - - = - - = - - = - v v v v v t m k v v dv v v From Eq. 5.9, , s 204 . 0 ) s m 8 . 9 ( ) s m 0 . 2 ( 2 2 t = = = g v k m and s 0.283 (1.386)
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Unformatted text preview: s) 204 . ( ) 386 . 1 ( = =--= k m t to the top. Equation 5.10 in the text gives us m 26 . ) 1 s)(e 204 . ( ) s m (2.0 s) (0.283 s) m . 2 ( ) 1 ( ) 1 ( 387 . 1 m) k ( t t m) k ( t t ) m k ( t t ) m k ( t =-+ =-+ =-=-=-=----- t x t t t t t e k m v t v dt e v dt v dx e v v e v dt dx...
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