# Drag forces on a skydiver

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1. Dec 28, 2015

### Will Freedic

1. The problem statement, all variables and given/known data
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

3. The attempt at a solution
I can find velocity as a function of time easy enough but not a function of distance

2. Dec 28, 2015

### BvU

Hi Will,

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. Dec 28, 2015

### SteamKing

Staff Emeritus
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. Dec 28, 2015

### haruspex

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. Dec 28, 2015

### Will Freedic

Thank you!, so is it correct to say that:

Cv2 = ma
cv2 = mv dvdx.
Cm/x = v2 / 2 + c
Cheers

6. Dec 28, 2015

### haruspex

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. Dec 29, 2015

### Will Freedic

Thank you for your help, i am new to these forums so am a little slow!, managed to get there now tho