To calculate the terminal velocity of the ball, we can use the formula for terminal velocity \( V_t \):
\[ V_t = \sqrt{\frac{2mg}{\rho C_d A}} \]
Where:
- \( m \) is the mass of the object (0.15 kg),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 , \text{m/s}^2 \)),
- \( \rho \) is the air density (1.225 kg/m³),
- \( C_d \) is the drag coefficient (0.007),
- \( A \) is the cross-sectional area (0.0026 m²).
Now, plug in the values:
- Calculate the numerator \( 2mg \):
\[ 2mg = 2 \cdot 0.15 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 2.943 , \text{N} \]
- Calculate the denominator \( \rho C_d A \):
\[ \rho C_d A = 1.225 , \text{kg/m}^3 \cdot 0.007 \cdot 0.0026 , \text{m}^2 = 2.2357 \times 10^{-5} , \text{kg/m} \]
- Now, substitute these values back into the formula for \( V_t \):
\[ V_t = \sqrt{\frac{2.943}{2.2357 \times 10^{-5}}} = \sqrt{131526.06} \approx 363 , \text{m/s} \]
Thus, the terminal velocity of the ball is approximately 363 m/s.
The correct response is:
363 m/s