To solve this problem, we need to break it down into two stages:

1. Free fall without the parachute (20 seconds): Here, the woman is falling freely, so we need to calculate her height and velocity at the end of this period.
2. Descent with the parachute open: Here, air resistance changes, and we need to calculate the remaining time it will take her to reach the ground.

The forces acting on the woman during both stages are her weight and air resistance. We will use the fact that air resistance is proportional to her velocity, with different constants $p$ for the two stages.

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Stage 1: Free Fall without Parachute

1. Governing Equation for Motion:

   The equation of motion, considering air resistance $pv$, is given by: $m \frac{dv}{dt} = mg - pv$ where:

   • $m$ is the mass of the woman,
   • $g = 32.2 \, \text{ft/s}^2$ is the acceleration due to gravity,
   • $p = 0.15 \, \text{lb-s/ft}$ is the resistance coefficient (without the parachute),
   • $v$ is the velocity.

   Rearranging the equation: $\frac{dv}{dt} = g - \frac{p}{m} v$

   The solution to this first-order differential equation is: $v(t) = \frac{mg}{p} \left(1 - e^{-\frac{p}{m}t} \right)$ where $v(0) = 0$.

2. Terminal Velocity: At $t = 20$ seconds (before the parachute opens), we can find the velocity $v(20)$ by solving the velocity equation above.

   The terminal velocity (if she falls long enough) is: $v_{\text{terminal}} = \frac{mg}{p}$

3. Height after 20 seconds: The position $y(t)$ is obtained by integrating the velocity equation: $y(t) = \int v(t) \, dt$ This will give us the height $y(20)$ at the time the parachute opens.

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Stage 2: Descent with the Parachute

Once the parachute opens, the air resistance coefficient changes to $p = 1.5 \, \text{lb-s/ft}$. The new equation of motion becomes: $m \frac{dv}{dt} = mg - 1.5v$ We follow the same steps as in Stage 1 to find the new velocity and time it takes for her to descend the remaining distance to the ground.

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Outline for Solution Steps

1. Determine the velocity after 20 seconds of free fall: $v(20) = \frac{mg}{0.15} \left(1 - e^{-\frac{0.15}{m} \times 20} \right)$

2. Calculate the height at 20 seconds: Use the position equation to find how far she has fallen in 20 seconds.

3. Determine the velocity and descent time with the parachute open: Use the same approach as in Stage 1, with the new air resistance coefficient $p = 1.5$.

4. Sum the total time: Add the time of free fall and the descent time after the parachute opens.

Would you like me to proceed with the full calculations for these steps?

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Here are some related questions that might help you deepen your understanding:

1. How does terminal velocity affect the time to reach the ground?
2. What would change if the air resistance coefficient were higher?
3. How is the time of descent affected if the parachute were opened earlier?
4. How would a different altitude (e.g., 15,000 ft) change the overall descent time?
5. How does varying the mass of the woman affect her terminal velocity?

Tip: Understanding the terminal velocity helps in estimating how fast an object will fall when the force of air resistance balances gravity.