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March 1, 2015

March 1, 2015

Posted by **Jacqueline** on Tuesday, October 14, 2008 at 11:51pm.

F(sub)d = bv^2, where b=0.14 N*s/m^2.

a) if he is falling w/ an unopened parachute at 64m/s, whats is the force of air resistance acting on him?

b) what is his acceleration?

c) what is his terminal velocity?

I figured that to be:

v(sub)t= sqrt (120 *9.8) / (0.14)

Can someone please help me find acceleration and the force of air resistance?

Thank you!! :)

- College Physics -
**drwls**, Wednesday, October 15, 2008 at 5:33ama) Use the formula that was provided. The drag force is

F = b*(64)^2 = 573 N at v = 64 m/s

b) a = F/Mass

c) Let the terminal velocity be V. It occurs when bV^2 = M g. That is before the parachute opens.

V = sqrt (M g/b)

which is what you had

The value of b will increase by a large factor when the parachute opens, but the question is not concerned with that.

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