Posted by Sydney on .
For beginner parachutists, the terminal velocity must be less than 5.00 m/s, and the parachutes used have a kvalue that is normally distributed with a mean of 2.05/s and a standard deviation of 0.04/s. If the acceleration due to gravity is 9.81 m/s^2, determine the probability, to the nearest thousandth, that the terminal velocity is less than 5.00 m/s.
Please help me, the help in the last question was greatly appreciated. Thankyou!

Physics 
Sydney,
Somebody please help me, I am really having problems. Thankyou!

Physics 
MathMate,
Could you please post the relevant equations and your attempt at the solution?
What is a kvalue for a parachute?
Thanks 
Physics 
Sydney,
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. 
Physics 
Damon,
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 
Physics 
MathMate,
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
= 2.051.962
=0.088
Standard deviation (meausure of variability of the kvalue)
= 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,
Thankyou, very much!!!