A 0.560 - kg ball has a terminal speed of \( 34.9 \mathrm{~m} / \mathrm{s} \). a) If the ball experiences a drag force of magnitude \( F_{D}=b v^{2} \), what is the value of the constant b? \[ b=\square \text { What is the SI unit of b? Give your answer in terms of base SI units. } \] Unit \( =\square \) b) What is the magnitude of the drag force when the speed of the baseball is \( 27.4 \mathrm{~m} / \mathrm{s} \) ? \( F_{D}=\square \)

To find the constant \( b \), we know at terminal velocity, the drag force equals the gravitational force. The gravitational force \( F_g \) acting on the ball can be calculated using the formula \( F_g = m \cdot g \), where \( m = 0.560 \, \text{kg} \) and \( g \approx 9.81 \, \text{m/s}^2 \). This gives us \( F_g \approx 5.49 \, \text{N} \). Using the drag force equation \( F_D = b v^2 \), we can set \( F_D = F_g \) at terminal velocity: \[ 5.49 = b(34.9)^2 \] Calculating \( (34.9)^2 \) gives approximately \( 1217.01 \), so: \[ b \approx \frac{5.49}{1217.01} \approx 0.00451 \, \text{kg/m} \] The SI unit of \( b \) can be derived from the drag force equation. The units of force are kilograms meter per second squared \((\text{kg} \cdot \text{m/s}^2)\), and \( v^2 \) has units of \((\text{m/s})^2\), which is \(\text{m}^2/\text{s}^2\). Therefore, the unit of \( b \) is: \[ \text{SI unit of } b = \frac{\text{N}}{\text{m}^2/\text{s}^2} = \frac{\text{kg} \cdot \text{m}/\text{s}^2}{\text{m}^2/\text{s}^2} = \text{kg/m} \] For part (b), we now need to calculate the drag force \( F_D \) when the speed \( v = 27.4 \, \text{m/s} \). Using the equation: \[ F_D = b v^2 \] Substituting \( b \approx 0.00451 \, \text{kg/m} \) and \( v = 27.4 \): \[ F_D = 0.00451 \cdot (27.4)^2 \] Calculating \( (27.4)^2 \) gives approximately \( 750.76 \): \[ F_D \approx 0.00451 \cdot 750.76 \approx 3.38 \, \text{N} \] So we arrive at the final answers: \[ b \approx 0.00451 \, \text{kg/m} \] \[ F_D \approx 3.38 \, \text{N} \]