Answer :
Explanation:
To determine the steady speed at which the buoy will rise to the surface of the ocean, we need to calculate the buoy's terminal velocity. Terminal velocity occurs when the buoy's weight is balanced by the buoyant force and the drag force due to the fluid (water) it is moving through.
Given data:
- Diameter of the buoy, \( D = 1.25 \) ft (radius \( r = \frac{D}{2} = 0.625 \) ft)
- Weight of the buoy, \( W = 33 \) lb
- Drag coefficient, \( C_D = 0.45 \)
First, calculate the cross-sectional area \( A \) of the buoy:
\[ A = \pi r^2 = \pi \left(0.625\right)^2 = \pi \times 0.390625 \approx 1.227 \text{ square feet} \]
Next, calculate the buoyant force \( F_b \):
\[ F_b = \rho V g \]
where
- \( \rho \) is the density of water (approximately \( 62.4 \) lb/ft\(^3\)),
- \( V \) is the volume of the buoy,
- \( g \) is the acceleration due to gravity (\( 32.2 \) ft/s\(^2\)).
The volume \( V \) of the buoy:
\[ V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left(0.625\right)^3 = \frac{4}{3} \pi \times 0.244140625 \approx 1.028 \text{ cubic feet} \]
Buoyant force \( F_b \):
\[ F_b = \rho V g = 62.4 \times 1.028 \times 32.2 \approx 2063 \text{ lb} \]
Now, calculate the terminal velocity \( v_t \) using the formula:
\[ v_t = \sqrt{\frac{2W}{\rho A C_D}} \]
Substitute the known values:
\[ v_t = \sqrt{\frac{2 \times 33}{62.4 \times 1.227 \times 0.45}} \]
\[ v_t = \sqrt{\frac{66}{34.0506}} \]
\[ v_t \approx \sqrt{1.937} \]
\[ v_t \approx 1.39 \text{ ft/s} \]
Therefore, the steady speed at which the buoy will rise to the surface of the ocean is approximately \( \boxed{1.39} \) ft/s.
