problem02_44 solution

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

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2.44: a) Using a y = –g, v 0 y = 5.00 s m and y 0 = 40.0 m in Eqs. (2.8) and (2.12) gives i) at t = 0.250 s, y = (40.0 m) + (5.00 s m )(0.250 s) – (1/2)(9.80 2 s m )(0.250 s) 2 = 40.9 m, v y = (5.00 s m ) – (9.80 2 s m )(0.250 s) = 2.55 s m and ii) at t = 1.00 s, y = (40.0 m) + (5.00 m/s)(1.00 s) – (1/2)(9.80 m/s 2 )(1.00 s) 2 = 40.1 m, v y = (5.00 s m ) – (9.80 2 s m )(1.00 s) = – 4.80 s m . b) Using the result derived in Example 2.8, the time is t = ) s m 80 . 9 ( ) m 0 . 40 0 )( s m 80 . 9 ( 2 ) s m 00 . 5 ( ) s m 00
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Unformatted text preview: ( 2 2 2--+ = 3.41 s. c) Either using the above time in Eq. (2.8) or avoiding the intermediate calculation by using Eq. (2.13), , s m 809 ) m . 40 )( s m 80 . 9 ( 2 ) s m 00 . 5 ( ) ( 2 2 2 2 2 2 2 =--=--= y y g v v y y v y = 28.4 s m . d) Using v y = 0 in Eq. (2.13) gives . m 2 . 41 m . 40 ) s m 80 . 9 ( 2 ) s m 00 . 5 ( 2 2 2 2 = + = + = y g v y e)...
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