ch05-p076 - first touch the patio until the moment his body...

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76. (a) For the 0.50 meter drop in “free-fall”, Eq. 2-16 yields a speed of 3.13 m/s. Using this as the “initial speed” for the final motion (over 0.02 meter) during which his motion slows at rate “ a ”, we find the magnitude of his average acceleration from when his feet
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Unformatted text preview: first touch the patio until the moment his body stops moving is a = 245 m/s 2 . (b) We apply Newtons second law: F stop mg = ma F stop = 20.4 kN....
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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.

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