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December 20, 2014

December 20, 2014

Posted by **yoyoma** on Tuesday, August 17, 2010 at 11:39pm.

- Physics -
**drwls**, Wednesday, August 18, 2010 at 12:07amDo you mean 255 m after the parachute opens?

The velocity when the chute opens is

V1 = sqrt(2aX) = 138.9 m/s

The velocity V2 after traveling a distance X' = 255 m is calculable with V2^2 - V1^2 = 2*a*X

where a = -6.1 m/s^2 and X = 255 m

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