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To find the y-component \( V_y \) of the given vector \( \vec{V} \), we can use the following formula for the y-component of a vector:

\[
V_y = V \cdot \sin(\theta)
\]

Where:
- \( V = 45.6 \, \text{m} \) is the magnitude of the vector.
- \( \theta = 109^\circ \) is the angle the vector makes with the positive x-axis.
- \( \sin(\theta) \) is the sine of the angle.

Now, we calculate:

\[
V_y = 45.6 \cdot \sin(109^\circ)
\]

Using the sine of \( 109^\circ \), which is approximately \( \sin(109^\circ) \approx 0.9455 \):

\[
V_y = 45.6 \cdot 0.9455 \approx 43.11 \, \text{m}
\]

Thus, the y-component of the vector is approximately:

\[
V_y \approx 43.11 \, \text{m}
\]

Would you like more details or have any questions about this calculation?

Here are 5 related questions for further exploration:
1. How would you calculate the x-component of this vector?
2. What would change if the angle \( \theta \) were smaller, such as \( 45^\circ \)?
3. How do you find the resultant magnitude and direction if both x and y components are known?
4. What happens if the vector is in a different quadrant?
5. How does the vector's angle affect its y-component's sign?

**Tip:** Remember that for angles greater than \( 90^\circ \), the sine function remains positive in the second quadrant, giving positive y-components.

Find the y-component of the vector: V = 45.6 m, θ = 109°

Learn how to find the y-component of a vector with a magnitude of 45.6 m and an angle of 109° using trigonometric principles. This problem demonstrates how to apply the sine rule to calculate the vertical component of the vector.

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