Somebody please help me, I am really having problems. Thank-you!
Could you please post the relevant equations and your attempt at the solution?
What is a k-value for a parachute?
Okay, relevant equations:
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.
so 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
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
You 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
here's some lighter reading:
and here's one on statistics:
It's the last one that concerns you most.
k=2.05 (mean value)
so the mean landing velocity is
The minimum value (kmin) of k to have a landing velocity of 5.00 m/s is
Difference from mean
Standard deviation (meausure of variability of the k-value)
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:
This is a quickie that does not really help you with your exams. Do your reading and prepare yourself accordings.
Thank-you, very much!!!
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