A skydiver weighing 200 lbs with clothes that have a drag coefficient of .325 is falling in an area that has an atmospheric density of 1.225 kg/m2 (and assuming that altitude has a negligible effect on atmospheric density). The skydiver can change the body orientation from straight-erect with a cross-sectional area of 6 sqft to a belly-flat cross-sectional area of 24 sqft. Calculate the terminal velocity of the person when the body has straight and when the body has belly-flat orientations. Calculate the terminal velocity on these two different orientations.
The Correct Answer and Explanation is:
To calculate the terminal velocity of a skydiver in two different orientations—straight-erect and belly-flat—we can use the formula for terminal velocity:
[
v_t = \sqrt{\frac{2mg}{\rho C_d A}}
]
Where:
- (v_t) = terminal velocity (m/s)
- (m) = mass of the skydiver (kg)
- (g) = acceleration due to gravity (9.81 m/s²)
- (\rho) = atmospheric density (kg/m³)
- (C_d) = drag coefficient (dimensionless)
- (A) = cross-sectional area (m²)
Step 1: Convert the weight and areas
- Weight (200 lbs):
- Convert weight to mass using the relation (1 \, \text{lb} = 0.453592 \, \text{kg}):
[
m = 200 \, \text{lbs} \times 0.453592 \, \text{kg/lb} \approx 90.72 \, \text{kg}
]
- Cross-Sectional Areas:
- Straight-erect: (6 \, \text{sqft} \to 6 \, \text{sqft} \times 0.092903 \, \text{m}^2/\text{sqft} \approx 0.557 \, \text{m}^2)
- Belly-flat: (24 \, \text{sqft} \to 24 \, \text{sqft} \times 0.092903 \, \text{m}^2/\text{sqft} \approx 2.229 \, \text{m}^2)
Step 2: Atmospheric Density
- Given atmospheric density: (\rho = 1.225 \, \text{kg/m}^3)
Step 3: Calculate Terminal Velocity for Each Orientation
Straight-Erect Orientation
Using (C_d = 0.325) and (A = 0.557 \, \text{m}^2):
[
v_t = \sqrt{\frac{2 \times 90.72 \, \text{kg} \times 9.81 \, \text{m/s}^2}{1.225 \, \text{kg/m}^3 \times 0.325 \times 0.557 \, \text{m}^2}}
]
Calculating the numerator:
[
2 \times 90.72 \times 9.81 \approx 1783.07
]
Calculating the denominator:
[
1.225 \times 0.325 \times 0.557 \approx 0.2130
]
Now calculating terminal velocity:
[
v_t \approx \sqrt{\frac{1783.07}{0.2130}} \approx \sqrt{8374.91} \approx 91.64 \, \text{m/s}
]
Belly-Flat Orientation
Using (A = 2.229 \, \text{m}^2):
[
v_t = \sqrt{\frac{2 \times 90.72 \times 9.81}{1.225 \times 0.325 \times 2.229}}
]
Calculating the new denominator:
[
1.225 \times 0.325 \times 2.229 \approx 0.8823
]
Now calculating terminal velocity:
[
v_t \approx \sqrt{\frac{1783.07}{0.8823}} \approx \sqrt{2016.88} \approx 44.87 \, \text{m/s}
]
Summary of Results
- Terminal Velocity (Straight-Erect): (\approx 91.64 \, \text{m/s})
- Terminal Velocity (Belly-Flat): (\approx 44.87 \, \text{m/s})
Explanation
The terminal velocity is the speed at which the force of gravity is balanced by the drag force acting on the falling object, resulting in zero acceleration. The formula highlights the dependence on several factors, including mass, cross-sectional area, drag coefficient, and atmospheric density.
In the straight-erect position, the skydiver’s smaller cross-sectional area (0.557 m²) allows for less air resistance compared to the belly-flat position (2.229 m²). Consequently, the terminal velocity is significantly higher when the body is straight-erect because the increased surface area in the belly-flat position increases drag, slowing the descent. This phenomenon underscores the importance of body orientation in freefall and illustrates how skydivers can control their fall speed through positioning.