1. **Problem statement:** Find the magnitude and direction of the vector sum \(\vec{A} + \vec{B}\) given vectors \(\vec{A}\) and \(\vec{B}\) from Figure E1.28. 2. **Given:** - \(\vec{A}\) has magnitude 8.00 m, points vertically downward (along negative y-axis). - \(\vec{B}\) has magnitude 15.0 m, points upward at 30° to the y-axis. 3. **Step 1: Express vectors in components.** - \(\vec{A} = 0\hat{i} - 8.00\hat{j}\) - For \(\vec{B}\), angle 30° from y-axis means 60° from x-axis (since y-axis is vertical): \[ B_x = 15.0 \sin 30^\circ = 15.0 \times 0.5 = 7.5 \] \[ B_y = 15.0 \cos 30^\circ = 15.0 \times 0.866 = 12.99 \] So, \(\vec{B} = 7.5\hat{i} + 12.99\hat{j}\) 4. **Step 2: Calculate \(\vec{A} + \vec{B}\) components:** \[ (A+B)_x = 0 + 7.5 = 7.5 \] \[ (A+B)_y = -8.00 + 12.99 = 4.99 \] 5. **Step 3: Find magnitude of \(\vec{A} + \vec{B}\):** \[ |\vec{A} + \vec{B}| = \sqrt{7.5^2 + 4.99^2} = \sqrt{56.25 + 24.90} = \sqrt{81.15} = 9.01 \text{ m} \] 6. **Step 4: Find direction (angle \(\theta\)) relative to positive x-axis:** \[ \theta = \tan^{-1} \left( \frac{4.99}{7.5} \right) = \tan^{-1}(0.665) = 33.7^\circ \] 7. **Answer:** The vector sum \(\vec{A} + \vec{B}\) has magnitude approximately 9.01 m and points at 33.7° above the positive x-axis. --- **Note:** Only the first question (part a) is solved as per instructions.