What Is the Terminal Velocity of a 70 kg Sky Diver?

[link1388]

Homework Statement

A sky diver with a mass of 70 kg jumps from an aircraft. The aerodynamic drag force acting on
the sky diver is known to be FD =kV^2

, where k=.25 N.s^2/m^2

Determine the maximum speed of free fall for the sky diver and the speed reached after 100 m of fall. Plot the speed of the sky
diver as a function of time and as a function of distance fallen

Homework Equations

F=ma

Fg - kV^2 = ma

The Attempt at a Solution

Well I did find Velocity Max to be

mg -.25V^2 = ma

mg - .25V^2 = 0 (a=0 because of terminal velocity)

Vmax = 52.4 m/s

But I am really confused on finding V(x) so I can find the speed at 100m? : /

 

[edgepflow]

First find V(t). Apply Newton's 2nd law again:

Fnet = ma = m dv/dt

The forces acting on the sky diver, as you noticed are:

Fnet = mg - k v^2

Now you can solve this differential equation for V(t).

The position function, s(t), can then be found from: V(t) = ds(t) / dt.

 

[RTW69]

Great problem. The first integration is pretty straight forward but the second integration to find dx/dt=v is ugly as is the simplifications. I would suggest going to the Wolframalpha.com site for the math.

 

[link1388]

Ok I think I got the first integration but the second ones definitely going to be tricky lol Thanks.