To find the terminal velocity of the ball, we can use the formula for terminal velocity (\(v_t\)) which is given by:
\[ v_t = \sqrt{\frac{2mg}{\rho C_d A}} \]
Where:
- \(m\) = mass of the ball (0.15 kg)
- \(g\) = acceleration due to gravity (approximately 9.81 m/s²)
- \(\rho\) = air density (1.225 kg/m³)
- \(C_d\) = drag coefficient (0.007)
- \(A\) = cross-sectional area of the ball (0.0026 m²)
Now, plug in the values:
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Calculate the weight (\(mg\)): \[ mg = 0.15 , \text{kg} \times 9.81 , \text{m/s}^2 = 1.4715 , \text{N} \]
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Substitute \(m\), \(g\), \(\rho\), \(C_d\), and \(A\) into the terminal velocity formula: \[ v_t = \sqrt{\frac{2 \times 1.4715 , \text{N}}{1.225 , \text{kg/m}^3 \times 0.007 \times 0.0026 , \text{m}^2}} \] Simplifying the denominator: \[ = 1.225 \times 0.007 \times 0.0026 = 2.2381 \times 10^{-5} , \text{kg/m}^2 \]
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Now calculate \(v_t\): \[ v_t = \sqrt{\frac{2 \times 1.4715}{2.2381 \times 10^{-5}}} \] \[ = \sqrt{\frac{2.943}{2.2381 \times 10^{-5}}} \] \[ = \sqrt{131484.4} \] \[ \approx 362.5 , \text{m/s} \]
Rounding this value, we have a terminal velocity of approximately 363 m/s.
Thus, the answer is 3. 363 m / s.