A volleyball is spiked so that it has an initial velocity of 15 m/s directed downward at an angle of `55^(@)` below the horizontal. What is the horizontal component of the ball's velocity when the opposing player fields the ball?

Options

A. 6.8 m/s
B. 8.6 m/s
C. 12 m/s
D. 18 m/s

Answer

To solve the problem, we need to find the horizontal component of the volleyball's initial velocity when it is spiked. The volleyball is given an initial velocity of 15 m/s directed downward at an angle of 55 degrees below the horizontal.

### Step-by-step Solution:

1. **Identify the Given Information:**
- Initial velocity \( v = 15 \, \text{m/s} \)
- Angle \( \theta = 55^\circ \) (below the horizontal)

2. **Understand the Components of Velocity:**
- The initial velocity can be broken down into two components: horizontal and vertical.
- The horizontal component \( v_x \) can be calculated using the cosine of the angle, since it is adjacent to the angle.

3. **Calculate the Horizontal Component:**
\[ v_x = v \cdot \cos(\theta) \]
Substituting the values:
\[ v_x = 15 \, \text{m/s} \cdot \cos(55^\circ) \]

4. **Calculate \( \cos(55^\circ) \):**
- Using a calculator, we find:
\[ \cos(55^\circ) \approx 0.5736 \]

5. **Substitute Back to Find \( v_x \):**
\[ v_x = 15 \, \text{m/s} \cdot 0.5736 \approx 8.604 \, \text{m/s} \]

6. **Round the Result:**
- Rounding to one decimal place, we get:
\[ v_x \approx 8.6 \, \text{m/s} \]

### Final Answer:
The horizontal component of the ball's velocity when the opposing player fields the ball is approximately **8.6 m/s**.

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