An acrobat is fired from a cannon at 59 miles per hour at an angle of \( 44^{\circ} \) 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 use the equations: \[ v_{x} = v \cdot \cos(\theta) \] \[ v_{y} = v \cdot \sin(\theta) \] Plugging in the values, we have: - \( v = 59 \) miles per hour - \( \theta = 44^{\circ} \) Calculating \( v_{x} \): \[ v_{x} = 59 \cdot \cos(44^{\circ}) \approx 59 \cdot 0.7193 \approx 42.41 \text{ miles per hour} \] Calculating \( v_{y} \): \[ v_{y} = 59 \cdot \sin(44^{\circ}) \approx 59 \cdot 0.6956 \approx 41.00 \text{ miles per hour} \] So, the final answer is: \[ \mathbf{v}=\left[v_{x}, v_{y}\right]=[ 42.41, 41.00 ] \]