problem03_19

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

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3.19: a) s m 0 . 18 9 . 36 sin ) s m 0 . 30 ( 0 = ° = y v ; solving Eq. (3.18) for t with 0 0 = y and m 0 . 10 = y gives s 99 2 s, 68 . 0 s m 80 . 9 ) m 0 . 10 )( s m 80 . 9 ( 2 ) s m 0 . 18 ( ) s m 0 . 18 ( 2 2 2 . t = - ± = b) The x -component of velocity will be s m 0 . 24 9 . 36 cos ) s m 0 . 30 ( = ° at all times. The y -component, obtained from Eq. (3.17), is s m 3 . 11 at the earlier time and
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Unformatted text preview: s m 3 . 11-at the later. c) The magnitude is the same, s m . 30 , but the direction is now ° 9 . 36 below the horizontal....
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• earlier time, horizontal. t=

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