• Support PF! Buy your school textbooks, materials and every day products Here!

Drag forces on a skydiver

  • #1

Homework Statement


A sky driver reaches his terminal velocity of 373 ms^-1 and deploys his parachute at a height of 1500m, the drag force =bv^2 and weigh can be ignored. Find an expression for velocity as a function of distance traveled from the point the parachute was deployed

The Attempt at a Solution


I can find velocity as a function of time easy enough but not a function of distance[/B]
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
12,881
2,965
Hi Will, :welcome:

Please read the guidelines, use the template, don't delete parts of the template, show your work so far and we'll help you from the point where you get stuck.
 
  • #3
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666

Homework Statement


A sky driver reaches his terminal velocity of 373 ms^-1 and deploys his parachute at a height of 1500m, the drag force =bv^2 and weigh can be ignored. Find an expression for velocity as a function of distance traveled from the point the parachute was deployed
A terminal velocity of 373 m/s is quite a feat. It's also greater than Mach 1, which makes it unlikely to occur. Are you sure a decimal point hasn't been misplaced?
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,550
4,980
There is a standard trick for eliminating time from an acceleration equation.
##a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}##.
Note that if we apply this to F=ma and integrate wrt x we get ##\int F.dx=\frac 12 mv^2##.
 
  • #5
Thank you!, so is it correct to say that:

Cv2 = ma
cv2 = mv dvdx.
Cm/x = v2 / 2 + c
Cheers
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,550
4,980
Thank you!, so is it correct to say that:

Cv2 = ma
cv2 = mv dvdx.
Cm/x = v2 / 2 + c
Cheers
A couple of problems there. First, what happened to gravity? Secondly, I don't understand what you did to get from cv2 = mv dvdx (presumably you meant mv dv/dx) to Cm/x = v2 / 2 + c
 
  • #7
A couple of problems there. First, what happened to gravity? Secondly, I don't understand what you did to get from cv2 = mv dvdx (presumably you meant mv dv/dx) to Cm/x = v2 / 2 + c
Thank you for your help, i am new to these forums so am a little slow!, managed to get there now tho
 

Related Threads for: Drag forces on a skydiver

  • Last Post
Replies
10
Views
6K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
20K
  • Last Post
Replies
3
Views
912
  • Last Post
Replies
6
Views
6K
Replies
5
Views
2K
Replies
7
Views
4K
  • Last Post
Replies
8
Views
2K
Top