An acrobat is fired from a cannon at 61 miles per hour at an angle of \( 32^{\circ} \) with respect to the horizontal. What are the horizontal and vertical components, \( v_{x} \) and \( v_{y} \), of the velocity vector \( \mathbf{v} \) ? \( \mathbf{v}=\left[v_{x}, v_{y}\right]=[ \) Ex:5.67

To find the horizontal and vertical components of the velocity vector, we can use trigonometric functions. The horizontal component \( v_{x} \) is calculated using the cosine function, while the vertical component \( v_{y} \) uses the sine function. 1. Calculate \( v_{x} \): \[ v_{x} = v \cdot \cos(\theta) = 61 \cdot \cos(32^{\circ}) \approx 61 \cdot 0.8480 \approx 51.7 \text{ miles per hour} \] 2. Calculate \( v_{y} \): \[ v_{y} = v \cdot \sin(\theta) = 61 \cdot \sin(32^{\circ}) \approx 61 \cdot 0.5299 \approx 32.3 \text{ miles per hour} \] Thus, the velocity vector \( \mathbf{v} \) is approximately \( \left[v_{x}, v_{y}\right] \approx [51.7, 32.3] \).