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July 29, 2016
Posted by **Sydney** on Monday, June 1, 2009 at 5:29pm.

Please help me, the help in the last question was greatly appreciated. Thank-you!

- Physics -
**Sydney**, Monday, June 1, 2009 at 6:30pmSomebody please help me, I am really having problems. Thank-you!

- Physics -
**MathMate**, Monday, June 1, 2009 at 6:39pmCould you please post the relevant equations and your attempt at the solution?

What is a k-value for a parachute?

Thanks - Physics -
**Sydney**, Monday, June 1, 2009 at 6:52pmOkay, relevant equations:

v(subscript)t=g/k

or

k=g/v(subscript)t

My attempt at the soloution, honestly, I have no clue, me and a friend have attempted this question and we still cannot figure it out. Please help me, I am totally lost. - Physics -
**Damon**, Monday, June 1, 2009 at 7:07pmso v mean = 9.81/2.05 = 4.785 m/s

k one sigma lower = 2.05 -0.04 = 2.01/s

velocity 1 sigma above mean = 9.81/2.01 = 4.881 /s

so

sigma of velocity = 4.881 - 4.785 = .095

now 5.00 is how many sigmas above mean?

5.00 - 4.785 = .215

and z = .215/.095 = 2.26 sigmas above mean (pretty improbable)

I only have a crude normal distribution table hgere but here are two values:

z = 2.2 then F(z) = .986

z = 2.3 then F(z) = .989

so we are talking around .988 - Physics -
**MathMate**, Monday, June 1, 2009 at 7:11pmYou and your mate would be better off reading something about terminal velocity. You cannot post questions to Jiskha in the exam hall! :)

Here are some background reading:

this one from NASA

http://exploration.grc.nasa.gov/education/rocket/termvr.html

here's some lighter reading:

http://www.northallertoncoll.org.uk/physics/module%202/terminal%20velocity/terminal%20velocity.htm

and here's one on statistics:

http://en.wikipedia.org/wiki/Standard_deviation

It's the last one that concerns you most.

k=g/vt

so

vt=g/k

g=9.81 m/s/s

k=2.05 (mean value)

so the mean landing velocity is

vt=9.81/2.05=4.785

The minimum value (kmin) of k to have a landing velocity of 5.00 m/s is

5=9.81/km

or

kmin=9.81/5=1.962

Difference from mean

= 2.05-1.962

=0.088

Standard deviation (meausure of variability of the k-value)

= 0.04

Therefore, a parachute has to have a value of k at 0.088/0.04=2.2 standard deviations below the mean value to have a landing velocity higher than 5 m/s.

If you look up a table of normal distribution for a tail end of 2.2 standard deviations (sigma), you will find that the probability is 0.0139, i.e. there is a 1.39% chance that the skydivers will land at more than 5 m/s.

The table is available here:

http://www.math.unb.ca/~knight/utility/NormTble.htm

This is a quickie that does not really help you with your exams. Do your reading and prepare yourself accordings.

Good luck. - Physics -
**Sydney**, Monday, June 1, 2009 at 7:23pmThank-you, very much!!!