To calculate the terminal velocity of the ball, we can use the formula:
\[ V_t = \sqrt{\frac{2mg}{\rho C_d A}} \]
where:
- \( V_t \) is the terminal velocity,
- \( m \) is the mass of the object (0.15 kg),
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s²),
- \( \rho \) is the density of the fluid (air) (1.225 kg/m³),
- \( C_d \) is the coefficient of drag (0.007),
- \( A \) is the cross-sectional area of the object (0.0026 m²).
Now, 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 values step-by-step:
-
Calculate the numerator: \[ 2 \times 0.15 \times 9.81 = 2.943 , \text{kg m/s}^2 \]
-
Calculate the denominator: \[ 1.225 \times 0.007 \times 0.0026 = 0.000022225 , \text{kg/(m s²)} \]
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Now plug these into the terminal velocity formula: \[ V_t = \sqrt{\frac{2.943}{0.000022225}} \approx \sqrt{132,434.56} \approx 364.2 , \text{m/s} \]
Rounding, we can say the terminal velocity is approximately \( 363 , \text{m/s} \).
Therefore, the correct choice is:
363 m/s
1 answer