To calculate the terminal velocity of the ball, we can use the formula for terminal velocity \( v_t \) for an object falling through a fluid:
\[ v_t = \sqrt{\frac{2mg}{\rho C_d A}} \]
Where:
- \( m \) is the mass of the ball (0.15 kg)
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s²)
- \( \rho \) is the air density (1.225 kg/m³)
- \( C_d \) is the drag coefficient (0.007)
- \( A \) is the surface area of the ball (0.0026 m²)
Now, let's substitute the values into the formula:
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Calculate \( 2mg \): \[ 2mg = 2 \times 0.15 , \text{kg} \times 9.81 , \text{m/s}^2 = 2.943 , \text{N} \]
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Calculate \( \rho C_d A \): \[ \rho C_d A = 1.225 , \text{kg/m}^3 \times 0.007 \times 0.0026 , \text{m}^2 = 0.000021155 , \text{kg/m} \]
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Now substitute these values into the terminal velocity formula: \[ v_t = \sqrt{\frac{2.943}{0.000021155}} = \sqrt{139,723.239} \approx 374.4 , \text{m/s} \]
However, this value does not match the options provided in your question, suggesting a discrepancy in the input parameters or the drag coefficient provided.
Considering the standard values for a ball-like object, you might want to double-check the coefficient of drag because a typical coefficient for a sphere in air can range from about 0.47 to higher values.
But based on provided parameters and calculation, the computed terminal velocity does not yield any of the options accurately. Therefore, if we trust the calculation performed based on conventional physics directly associated with falling objects, and if the values appear to be accurate, the response might not match the available answers.
Still, if I had to choose the best option based on the calculation performed, I would select the closest value to the computed result. The closest, of your provided options, is 363 m/s.