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The problem asks to determine the horizontal component of vector \( \vec{V} \) with a magnitude of \( 91.5 \, \text{m/s} \) and an angle of \( 78.7^\circ \) above the horizontal.

To calculate the horizontal component \( V_x \), we use the cosine function:
\[
V_x = V \cdot \cos(\theta)
\]
Where:
- \( V = 91.5 \, \text{m/s} \) is the magnitude of the vector,
- \( \theta = 78.7^\circ \) is the angle from the horizontal.

Let's calculate the horizontal component.The horizontal component of vector \( \vec{V} \) is approximately \( 17.93 \). 

Do you want more details or have any questions?

Here are 5 related questions for practice:
1. How would you calculate the vertical component of vector \( \vec{V} \)?
2. What is the significance of the cosine function when determining vector components?
3. If the angle were larger than \( 90^\circ \), how would it affect the horizontal component?
4. How can the magnitude of a vector be found from its components?
5. What is the relationship between vectors and coordinate systems in physics?

**Tip:** When working with vectors, always make sure the angle is measured from the horizontal (or the axis in question) to apply trigonometric functions correctly.

Determine the horizontal component of vector V. Include a (-) sign if necessary. Do not include any units.

Learn how to calculate the horizontal component of a vector with a magnitude of 91.5 m/s and an angle of 78.7°. This problem uses key concepts from trigonometry and vector decomposition, suitable for high school students in grades 9-12.

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