problem03_23

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

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3.23: a) With 0 = y v in Eq. (3.17), solving for t and substituting into Eq. (3.18) gives m 6 . 13 ) m/s 80 . 9 ( 2 0 . 33 sin ) m/s 0 . 30 ( 2 sin 2 ) ( 2 2 2 0 2 2 0 2 0 0 = ° = = = - g v g v y y y α b) Rather than solving a quadratic, the above height may be used to find the time the rock takes to fall from its greatest height to the ground, and hence the vertical component of velocity, m/s 7 . 23 ) m/s m)(9.80 6 . 28 ( 2 2 2 = = = yg v y , and so the speed of the rock is m/s 6 . 34 )) 33.0 m/s)(cos 0 . 30 (( m/s) 7 . 23
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Unformatted text preview: 2 2 = ° + . c) The time the rock is in the air is given by the change in the vertical component of velocity divided by the acceleration –g ; the distance is the constant horizontal component of velocity multiplied by this time, or m. 103 ) m/s 80 . 9 ( )) m/s)sin33. ((30.0 m/s 23.7 ( m/s)cos33. . 30 ( 2 =-°--° = x d)...
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