Here's the problem:
A skydiver and her equipment together weigh 192 pounds. Before the parachute is opened, there is an air drag force equal in magnitude to six times her velocity. Four seconds after stepping from the plane, the skydiver opens the parachute, producing a drag equal to three times the square of the velocity and how far the skydiver has fallen at time t. What is the terminal velocity?
I found this problem from Peter V. O'Neil's "Advanced Engineering Mathematics, 7E".
I already have answers for the velocity and distance covered when the parachute is not yet opened. I also know the skydiver's terminal velocity.
I need help in solving for the velocity and displacement functions once the parachute is opened.
Can anyone show me his/her solution for the entire problem?
Hint: You need to evaluate
$$ \int \frac{dv }{g-Kv^2} $$
Where $K=3/M$ in your specific problem. You can "cheat" by using wolfram or evaluate it yourself using partial fractions. The initial condition is the velocity at the end of the 4 second free-fall.