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Unformatted text preview: launched from the ground at t = with speed 9.35 m/s (see above) and ask of its height and speed at t = 0.200 s, using Eq. 215 and Eq. 211: y = (9.35 m/s)(0.200 s) – 1 2 (9.80 m/s 2 )(0.200 s) 2 = 1.67 m, v = 9.35 m/s – (9.80 m/s 2 )(0.200 s) = 7.39 m/s. The kinetic energy is K = 1 2 (2.50 kg) (7.39 m/s) 2 = 68.2 J. (d) The gravitational potential energy U = mgy = (2.5 kg)(9.8 m/s 2 )(1.67 m) = 41.0 J...
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Energy, Kinetic Energy

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