problem03_58

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

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3.58: Equation 3.27 relates the vertical and horizontal components of position for a given set of initial values. a) Solving for 0 v gives . tan cos 2 / 0 0 2 2 2 0 y x gx v - = α Insertion of numerical values gives m/s 6 . 16 0 = v . b) Eliminating t between Equations 3.20 and 3.23 gives y v as a function of x , . cos sin 0 0 0 0 v gx v v y - = Using the given values yields m/s, 98 . 6 m/s,
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Unformatted text preview: 28 . 8 cos-= = = y x v v v so m/s, 8 . 10 ) m/s 98 . 6 ( ) m/s 28 . 8 ( 2 2 =-+ = v at an angle of ( 29 -=-1 . 40 arctan 24 . 8 98 . 6 , with the negative sign indicating a direction below the horizontal. c) The graph of ) ( t v x is a horizontal line....
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