What is the magnitude of the acceleration of a skydiver at the instant she is falling at one-half her terminal speed?

To calculate the magnitude of the acceleration of a skydiver when she is falling at half her terminal speed, we need to analyze the forces involved. The two primary forces acting on a skydiver during free fall are:

1. **Gravitational Force (Weight)**: \( F_g = mg \)
2. **Drag Force (Air Resistance)**: \( F_d = \frac{1}{2} C_d \rho A v^2 \), where:
   - \( C_d \) is the drag coefficient,
   - \( \rho \) is the air density,
   - \( A \) is the cross-sectional area,
   - \( v \) is the velocity of the skydiver.

At terminal velocity, the gravitational force is balanced by the drag force:

\[
mg = \frac{1}{2} C_d \rho A v_T^2
\]

Where \( v_T \) is the terminal velocity. When the skydiver is falling at half her terminal velocity, the drag force will not fully counteract gravity.

### Step-by-step solution:

1. **At terminal velocity**: The net force is zero because the gravitational force is balanced by the drag force:
   \[
   F_g = F_d \quad \text{or} \quad mg = \frac{1}{2} C_d \rho A v_T^2
   \]

2. **At half terminal velocity**: The drag force is proportional to the square of the velocity. Therefore, when the skydiver is falling at \( \frac{v_T}{2} \), the drag force becomes:
   \[
   F_d = \frac{1}{2} C_d \rho A \left( \frac{v_T}{2} \right)^2 = \frac{1}{4} \left( \frac{1}{2} C_d \rho A v_T^2 \right) = \frac{1}{4} mg
   \]

   The drag force at half the terminal speed is one-fourth of the gravitational force.

3. **Net force at half terminal speed**: The net force acting on the skydiver is the difference between the gravitational force and the drag force:
   \[
   F_{\text{net}} = mg - \frac{1}{4} mg = \frac{3}{4} mg
   \]

4. **Acceleration at half terminal speed**: Using Newton's second law \( F = ma \), the acceleration is:
   \[
   a = \frac{F_{\text{net}}}{m} = \frac{\frac{3}{4} mg}{m} = \frac{3}{4} g
   \]

Thus, the magnitude of the acceleration of the skydiver at the instant she is falling at one-half her terminal speed is:

\[
a = \frac{3}{4} g
\]

### Final Answer:
The acceleration is \( \frac{3}{4} \times 9.8 \, \text{m/s}^2 = 7.35 \, \text{m/s}^2 \).

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Would you like any further details or clarifications on this solution?

Here are five related questions you might find interesting:
1. What factors influence the terminal velocity of a skydiver?
2. How does the drag coefficient \( C_d \) vary with the shape and orientation of the skydiver?
3. What is the terminal velocity for a typical skydiver in different positions (e.g., belly-to-earth, head-down)?
4. How does air density impact the terminal velocity in high-altitude jumps?
5. What happens to a skydiver's acceleration as they approach terminal velocity?

**Tip**: In real-world scenarios, terminal velocity depends on body posture, clothing, and the altitude, since air density decreases with altitude!

Learn how to calculate the acceleration of a skydiver when falling at half her terminal speed. This physics problem explores forces in free fall, including gravitational force and drag force, and applies Newton's second law. Suitable for students in Grades 9-12.

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