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