1. The problem statement, all variables and given/known data 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? 2. Relevant 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) 3. 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?