To find the terminal velocity of a falling ball, we can use the following formula for terminal velocity \(v_t\):
\[ v_t = \sqrt{\frac{2mg}{\rho C_d A}} \]
where:
- \(m\) = mass of the object (kg)
- \(g\) = acceleration due to gravity (\(9.81 , m/s^2\))
- \(\rho\) = air density (\(kg/m^3\))
- \(C_d\) = coefficient of drag
- \(A\) = surface area of the object (\(m^2\))
Now, substituting the values provided:
- \(m = 0.15 , kg\)
- \(g = 9.81 , m/s^2\)
- \(\rho = 1.225 , kg/m^3\)
- \(C_d = 0.007\)
- \(A = 0.0026 , m^2\)
Plugging in the values:
\[ v_t = \sqrt{\frac{2 \times 0.15 \times 9.81}{1.225 \times 0.007 \times 0.0026}} \]
Calculating the numerator:
\[ 2 \times 0.15 \times 9.81 = 2.943 \]
Calculating the denominator:
\[ 1.225 \times 0.007 \times 0.0026 \approx 0.000022573 \]
Now calculating \(v_t\):
\[ v_t = \sqrt{\frac{2.943}{0.000022573}} \approx \sqrt{130175.8678} \approx 360.5 , m/s \]
Rounding gives us approximately \(363 , m/s\).
Thus, the terminal velocity is \(\boxed{363 , m/s}\).