How Does a Skydiver's Velocity Change with Distance After Parachute Deployment?

Click For Summary

Homework Help Overview

The discussion revolves around the physics of a skydiver deploying a parachute and how their velocity changes with distance after deployment. The problem involves concepts of terminal velocity, drag force, and the relationship between velocity, distance, and acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the challenge of expressing velocity as a function of distance rather than time. Some question the validity of the given terminal velocity, while others suggest using a standard method to eliminate time from the equations of motion.

Discussion Status

The conversation is ongoing, with participants providing guidance on mathematical approaches and questioning assumptions made in the original problem statement. There is a recognition of potential errors in the initial conditions, particularly regarding the terminal velocity value.

Contextual Notes

Participants note the absence of gravity in some equations and express confusion over certain mathematical transitions. There is an emphasis on adhering to forum guidelines and providing clear work for review.

Will Freedic
Messages
6
Reaction score
0

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]
 
Physics news on Phys.org
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.
 
Will Freedic said:

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?
 
  • Like
Likes   Reactions: Will Freedic and BvU
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##.
 
Thank you!, so is it correct to say that:

Cv2 = ma
cv2 = mv dvdx.
Cm/x = v2 / 2 + c
Cheers
 
Will Freedic said:
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
 
haruspex said:
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
 

Similar threads

Replies
10
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
41
Views
5K
Replies
5
Views
10K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 10 ·
Replies
10
Views
8K
  • · Replies 14 ·
Replies
14
Views
2K