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We are given the height function \( h(t) = -4.9t^2 + 50 \), which describes the height of a ball dropped from a platform 50 meters high. The problem asks for the following:

### Part (a): Average Rate of Change of Height
We are tasked to compute the average rate of change of the height over different intervals around \( t = 2 \). The average rate of change of a function \( h(t) \) over an interval \( [t_1, t_2] \) is given by:

\[
\text{Average Rate of Change} = \frac{h(t_2) - h(t_1)}{t_2 - t_1}
\]

Let's compute this for each interval provided:

1. **Interval [2, 2.5]:**
   \[
   \text{Average Rate of Change} = \frac{h(2.5) - h(2)}{2.5 - 2}
   \]

2. **Interval [2, 2.25]:**
   \[
   \text{Average Rate of Change} = \frac{h(2.25) - h(2)}{2.25 - 2}
   \]

3. **Interval [2, 2.05]:**
   \[
   \text{Average Rate of Change} = \frac{h(2.05) - h(2)}{2.05 - 2}
   \]

4. **Interval [2, 2.001]:**
   \[
   \text{Average Rate of Change} = \frac{h(2.001) - h(2)}{2.001 - 2}
   \]

### Part (b): Instantaneous Rate of Change (IROC)
Based on the results from part (a), we suspect the IROC at \( t = 2 \), which is the derivative of \( h(t) \) at \( t = 2 \). The derivative of \( h(t) = -4.9t^2 + 50 \) is:

\[
h'(t) = -9.8t
\]

At \( t = 2 \):
\[
h'(2) = -9.8 \times 2 = -19.6
\]

This value represents the instantaneous rate of change (IROC) of the height at \( t = 2 \).

Let me calculate the values for part (a).Here are the average rates of change of the height on the specified intervals:

1. On the interval \([2, 2.5]\): \(-22.05\) m/s
2. On the interval \([2, 2.25]\): \(-20.83\) m/s
3. On the interval \([2, 2.05]\): \(-19.85\) m/s
4. On the interval \([2, 2.001]\): \(-19.60\) m/s

### Part (b): Instantaneous Rate of Change (IROC)
As suspected, based on these values, the IROC at \( t = 2 \) is approximately \(-19.6\) m/s, matching the derivative calculation.

Would you like further details or clarifications?

Here are 5 related questions to explore:
1. How would the average rate of change differ for longer intervals?
2. Can the height function be used to find when the ball hits the ground?
3. What does the IROC physically represent in the context of the problem?
4. How does the IROC relate to the velocity of the ball?
5. How would air resistance affect the rates of change?

**Tip:** When computing the IROC, you're essentially finding the derivative of the function, which gives the exact rate at a specific point.

The height of a ball dropped from a platform 50 meters off the ground, with initial velocity 0, is given by h(t) = -4.9t^2 + 50. Calculate the average rate of change of the height over intervals [2, 2.5], [2, 2.25], [2, 2.05], and [2, 2.001]. Based on these results, what is the Instantaneous Rate of Change (IROC) at t = 2?

Learn how to calculate the average rate of change of a quadratic height function over different intervals and determine the Instantaneous Rate of Change (IROC) at t = 2. The problem involves quadratic equations and introductory calculus concepts.

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