An orange falling from 28 feet has a height of
s(t) = 28 − 16t2 feet
when it has fallen for t seconds. 
Make a table to record the average rates that the orange is falling during the last second, half-second, quarter-second, and eighth-of-a-second of its fall. (Round your answers to three decimal places.) 

The height of the orange as a function of time is given by:

\[
s(t) = 28 - 16t^2 \, \text{feet}
\]

To calculate the **average rate of change** of the height (which represents the average velocity) during the last second, half-second, quarter-second, and eighth-of-a-second of its fall, we need to compute the difference in height over the difference in time, i.e.,

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

The orange reaches the ground when \(s(t) = 0\), so we first solve for \(t\) by setting \(s(t) = 0\):

\[
28 - 16t^2 = 0
\]
\[
t^2 = \frac{28}{16} = 1.75
\]
\[
t = \sqrt{1.75} \approx 1.323 \, \text{seconds}
\]

Thus, the orange hits the ground at \(t \approx 1.323\) seconds.

Now, we'll compute the average rate of change during different intervals before it hits the ground.

### 1. Last Second: \([t = 0.323 \text{ to } t = 1.323]\)

\[
\text{AROC} = \frac{s(1.323) - s(0.323)}{1.323 - 0.323}
\]

### 2. Last Half-Second: \([t = 0.823 \text{ to } t = 1.323]\)

\[
\text{AROC} = \frac{s(1.323) - s(0.823)}{1.323 - 0.823}
\]

### 3. Last Quarter-Second: \([t = 1.073 \text{ to } t = 1.323]\)

\[
\text{AROC} = \frac{s(1.323) - s(1.073)}{1.323 - 1.073}
\]

### 4. Last Eighth-of-a-Second: \([t = 1.198 \text{ to } t = 1.323]\)

\[
\text{AROC} = \frac{s(1.323) - s(1.198)}{1.323 - 1.198}
\]

Let's compute these values and round the answers to three decimal places.The average rates at which the orange is falling during the last second, half-second, quarter-second, and eighth-of-a-second of its fall are:

1. **Last second**: \(-26.332 \, \text{feet per second}\)
2. **Last half-second**: \(-34.332 \, \text{feet per second}\)
3. **Last quarter-second**: \(-38.332 \, \text{feet per second}\)
4. **Last eighth-of-a-second**: \(-40.332 \, \text{feet per second}\)

Would you like more details on these calculations or have any other questions?

Here are five follow-up questions:
1. How would the rates change if the height function had different coefficients?
2. What is the instantaneous rate of change at the time the orange hits the ground?
3. How would the average rate of change vary with other fractions of a second?
4. What is the maximum height the orange reaches before falling?
5. How long would it take for the orange to hit the ground if dropped from a different height?

**Tip**: The average rate of change is an approximation of velocity, but as the time intervals get smaller, it approaches the instantaneous velocity!

An orange falling from 28 feet has a height of s(t) = 28 − 16t^2 feet when it has fallen for t seconds. Make a table to record the average rates that the orange is falling during the last second, half-second, quarter-second, and eighth-of-a-second of its fall. (Round your answers to three decimal places.)

Explore how to calculate the average rate of change for an orange falling from 28 feet using s(t) = 28 − 16t^2. The problem involves determining the average velocity over different intervals, including the last second, half-second, quarter-second, and eighth-of-a-second. Suitable for students in Grades 11-12.

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