Free Falling (dynamics question)

[Raging Dragon]

Homework Statement

A sky diver jumps off a plane that is at 9570.7m above the earth. Assuming a combination of stokes and Newtonian drag, where Fdrag=-m*b*v-m*b^2*v^2 and assume constant g=9.81, at what point is the KE of the skydiver equal to the energy dissapated by the drag force?

Homework Equations

Equating mg=Fdrag one can find b quite easily.

I've derived through solving a nonlinear ODE Fnet=m*vdot=m*g-m*b*v-m*b^2*v^ 2 that

v(t)=-10.10573478+64.10573477*tanh(.1569282379*t+.1589673340)

The Attempt at a Solution

So I know that Gravitational Potential energy is =m*g*h=m*9.81*9570.7

Then I figure that PE=GPE (initially) for max h.

then when the sky diver jumps that PE=GPE+KE+Drag Energy

Now my question is to find the drag energy, do I integrate F_drag with respect to v, or t?

Then I take it that you need to also determine the time and distance fallen for this to happen, so then you goet PE-GPE=KE+Drag Energy and then KE=Drag Energy=(PE-GPE)/2

This make sense?

 

[OlderDan]

The energy dissipated by the drag force is the negative of the (negative) work it does on the diver, which is the integral of the force over the distance the diver falls.

 

[Raging Dragon]

Thanks for explaining what the boundary of the integral is, but against which variable if I have force as a function of velocity, and I have velocity as a function time?