Posted by Sydney on Monday, June 1, 2009 at 5:29pm.
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?
Thanks
Okay, 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.
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
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
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
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.
Thank-you, very much!!!