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The question asks for the instantaneous rate of change in the height of a soccer ball based on given time intervals. Here's how we can approach each part:

**Given Table:**

| Time (s)  | 0.0  | 0.5  | 1.0  | 1.5  | 2.0  | 2.5  | 3.0  | 3.5  | 4.0  | 4.5  | 5.0  |
|-----------|------|------|------|------|------|------|------|------|------|------|------|
| Height (m)| 0.5  | 11.78| 20.6 | 26.98| 30.9 | 32.38| 31.4 | 27.98| 22.1 | 13.78| 3.0  |

### (a) Preceding and Following Interval Method

To estimate the instantaneous rate of change at \( t = 2.0 \) using the preceding and following intervals, we can calculate the average rate of change using points just before and after \( t = 2.0 \).

1. **Preceding Interval (between 1.5s and 2.0s):**

   \[
   \text{Rate of change} = \frac{\text{Height at } t=2.0 - \text{Height at } t=1.5}{2.0 - 1.5}
   = \frac{30.9 - 26.98}{2.0 - 1.5} = \frac{3.92}{0.5} = 7.84 \text{ m/s}
   \]

2. **Following Interval (between 2.0s and 2.5s):**

   \[
   \text{Rate of change} = \frac{\text{Height at } t=2.5 - \text{Height at } t=2.0}{2.5 - 2.0}
   = \frac{32.38 - 30.9}{2.5 - 2.0} = \frac{1.48}{0.5} = 2.96 \text{ m/s}
   \]

Thus, using the preceding and following interval method, we get two rates:
- Preceding interval rate: 7.84 m/s
- Following interval rate: 2.96 m/s

### (b) Centered Interval Method

For the centered interval method, we use the points before and after \( t = 2.0 \), i.e., \( t = 1.5 \) and \( t = 2.5 \):

\[
\text{Rate of change} = \frac{\text{Height at } t=2.5 - \text{Height at } t=1.5}{2.5 - 1.5}
= \frac{32.38 - 26.98}{2.5 - 1.5} = \frac{5.4}{1.0} = 5.4 \text{ m/s}
\]

### (c) Preference of Method

The centered interval method typically provides a better estimate of the instantaneous rate of change because it takes into account values from both sides of the point in question. The preceding and following interval method only considers either the past or the future point, leading to a less balanced estimate.

Would you like more details or have any questions on this?

### Related Questions:
1. How do you interpret the instantaneous rate of change graphically?
2. Can you explain how to calculate the average rate of change over a broader interval?
3. Why does the centered interval method generally give a better approximation of the instantaneous rate of change?
4. How does the rate of change vary when the ball reaches its maximum height?
5. How would you calculate the acceleration of the ball at \( t = 2.0 \)?

### Tip:
When estimating instantaneous rates of change, using smaller intervals or the centered difference method can give more accurate results.

A soccer ball is kicked into the air. The following table of values shows the height of the ball above the ground at various times during its flight. Estimate the instantaneous rate of change in the height of the ball at exactly t = 2.0 s using different methods.

Explore how to estimate the instantaneous rate of change in the height of a soccer ball kicked into the air at t = 2.0 seconds using both the preceding and following interval method and the centered interval method. This problem is an excellent example of applying calculus concepts to real-world scenarios.

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