To find the terminal velocity of the skydiver, we can use the following formula for terminal velocity $v_t$:

$v_t = \sqrt{\frac{2mg}{\rho A C_d}}$

Where:

• $m$ = mass of the skydiver = 64.0 kg
• $g$ = acceleration due to gravity = 9.81 m/s²
• $\rho$ = air density = 1.21 kg/m³
• $A$ = cross-sectional area = 0.71 m²
• $C_d$ = drag coefficient = 1.0

Step-by-step calculation:

1. Gravitational Force: The weight of the skydiver is $mg$, where: $mg = 64.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 627.84 \, \text{N}$

2. Drag Force Formula: At terminal velocity, the gravitational force is balanced by the drag force: $F_{\text{drag}} = \frac{1}{2} \rho v_t^2 A C_d$ Set $F_{\text{drag}} = mg$, so: $627.84 = \frac{1}{2} (1.21) v_t^2 (0.71) (1.0)$

3. Solve for $v_t$: $627.84 = 0.42935 v_t^2$ $v_t^2 = \frac{627.84}{0.42935} = 1463.88$ $v_t = \sqrt{1463.88} = 38.3 \, \text{m/s}$

Thus, the terminal velocity is approximately 38 m/s.

Answer: 38 m/s

Let me know if you want further details or have any other questions!

Here are some related questions for you to explore:

1. How does air density affect terminal velocity?
2. What happens to the terminal velocity if the drag coefficient increases?
3. How does the mass of an object impact its terminal velocity?
4. What is the role of cross-sectional area in the drag equation?
5. How does terminal velocity change with altitude?

Tip: Terminal velocity is reached when the force of gravity pulling the object down equals the air resistance pushing it up.