problem02_43 solution

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

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2.43: a) Using the method of Example 2.8, the time the ring is in the air is ) s m 80 . 9 ( ) m 0 . 12 ( ) s m 80 . 9 ( 2 s) m 00 . 5 ( s) m 00 . 5 ( ) ( 2 2 2 2 0 2 0 0 - - + = - - + = g y y g v v t y y s, 156 . 2 = keeping an extra significant figure. The average velocity is then s m 57 . 5 s 2.156 m 0 . 12 = , down. As an alternative to using the quadratic formula, the speed of the ring when it hits the ground may be obtained from ) ( 2 0 2 0 2 y y g v v y y - - = , and the average velocity found from 2 0 y y v v + ; this is algebraically identical to the result obtained by the quadratic formula.
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Unformatted text preview: b) While the ring is in free fall, the average acceleration is the constant acceleration due to gravity, 2 s / m 80 . 9 down. c) 2 2 1 gt t v y y y-+ = 2 2 ) s m 8 . 9 ( 2 1 ) s m (5.00 m . 12 t t-+ = Solve this quadratic as in part a) to obtain t = 2.156 s. d) ) m . 12 )( s m 8 . 9 ( 2 ) s m (5.00 ) ( 2 2 2 2 2--=--= y y g v v y y s m 1 . 16 = y v e)...
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  • Acceleration, Velocity, extra significant figure

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