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A 75.0- man steps off a platform 3.10 above the ground. He keeps his legs straight as he falls, but at the moment his feet touch the ground his knees begin to bend, and, treated as a particle, he moves an additional 0.60 before coming to rest.

2007-09-26 11:11:39 · 1 answers · asked by dymestatus 1 in Science & Mathematics Physics

1 answers

I'm guessing that you mean 3.1 meters above the ground.
v = √(2gx) where v is m/s, g is 9.8 m/s² and x is distance, so
v = √(2 * 9.8 * 3.1) = 7.919 w/s

Doug

2007-09-26 11:19:57 · answer #1 · answered by doug_donaghue 7 · 0 2

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