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April 20, 2014

April 20, 2014

Posted by **demi** on Monday, August 30, 2010 at 7:36pm.

- physics -
**drwls**, Monday, August 30, 2010 at 7:49pmFirst compute the maximum speed using

V(max) = sqrt(2aX)

a is the acceleration and x = 402 m.

After the chute opens (which I will call t=0), the distance moved is

X = Vmax*t - 2.6 t^2

Solve for t when X = 375 m.

Then use that t to get the velocity at that time.

V(t) = Vmax - 5.20 t

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