To calculate the terminal velocity of the ball, you can use the following formula:
\[ v_t = \sqrt{\frac{2mg}{\rho C_d A}} \]
Where:
- \( v_t \) is the terminal velocity (m/s)
- \( m \) is the mass of the object (kg)
- \( g \) is the acceleration due to gravity (approximately \( 9.81 , \text{m/s}^2 \))
- \( \rho \) is the air density (kg/m³)
- \( C_d \) is the drag coefficient (dimensionless)
- \( A \) is the cross-sectional area (m²)
Given:
- \( m = 0.15 , \text{kg} \)
- \( g = 9.81 , \text{m/s}^2 \)
- \( \rho = 1.225 , \text{kg/m}^3 \)
- \( C_d = 0.007 \)
- \( A = 0.0026 , \text{m}^2 \)
Substituting the values into the formula:
\[ v_t = \sqrt{\frac{2 \times 0.15 , \text{kg} \times 9.81 , \text{m/s}^2}{1.225 , \text{kg/m}^3 \times 0.007 \times 0.0026 , \text{m}^2}} \]
Calculating the numerator:
\[ 2 \times 0.15 , \text{kg} \times 9.81 , \text{m/s}^2 = 2.9415 , \text{kg m/s}^2 \]
Calculating the denominator:
\[ 1.225 , \text{kg/m}^3 \times 0.007 \times 0.0026 , \text{m}^2 = 0.000022365 , \text{kg/m} \]
Now, substituting back to calculate \( v_t \):
\[ v_t = \sqrt{\frac{2.9415}{0.000022365}} \approx \sqrt{131,011.167} \]
Calculating the final result:
\[ v_t \approx 362.11 , \text{m/s} \]
Based on this computation, rounding to the nearest whole number gives:
Terminal velocity \( v_t \) is approximately 363 m/s.
Therefore, the correct response is 363 m/s.