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.