The height of a ball dropped from a platform 50 meters off the ground, with initial
velocity 0, at time t seconds, is given by: h(t) = −4.9t^2 + 50. 
24. a. Calculate the average rate of change of the height with respect to time on the intervals:
[2, 2.5], [2, 2.25], [2, 2.05], [2, 2.001].
b. Based on part (a), what do you suspect is the IROC of the height with respect to
time at t = 2?

To solve this problem, we will calculate the average rate of change (AROC) of the height of the ball over each of the given intervals and then estimate the instantaneous rate of change (IROC) of the height with respect to time at \( t = 2 \) seconds.

### Part (a): Calculating the Average Rate of Change

The formula for the average rate of change of a function \( h(t) \) over the interval \([a, b]\) is:

\[
\text{AROC} = \frac{h(b) - h(a)}{b - a}
\]

Given that the height function is:

\[
h(t) = -4.9t^2 + 50
\]

We will calculate the AROC for the intervals \([2, 2.5]\), \([2, 2.25]\), \([2, 2.05]\), and \([2, 2.001]\).

#### Interval \([2, 2.5]\)

\[
a = 2, \, b = 2.5
\]

\[
h(2) = -4.9(2^2) + 50 = -4.9(4) + 50 = -19.6 + 50 = 30.4
\]

\[
h(2.5) = -4.9(2.5^2) + 50 = -4.9(6.25) + 50 = -30.625 + 50 = 19.375
\]

\[
\text{AROC} = \frac{h(2.5) - h(2)}{2.5 - 2} = \frac{19.375 - 30.4}{0.5} = \frac{-11.025}{0.5} = -22.05
\]

#### Interval \([2, 2.25]\)

\[
a = 2, \, b = 2.25
\]

\[
h(2.25) = -4.9(2.25^2) + 50 = -4.9(5.0625) + 50 = -24.80625 + 50 = 25.19375
\]

\[
\text{AROC} = \frac{h(2.25) - h(2)}{2.25 - 2} = \frac{25.19375 - 30.4}{0.25} = \frac{-5.20625}{0.25} = -20.825
\]

#### Interval \([2, 2.05]\)

\[
a = 2, \, b = 2.05
\]

\[
h(2.05) = -4.9(2.05^2) + 50 = -4.9(4.2025) + 50 = -20.59125 + 50 = 29.40875
\]

\[
\text{AROC} = \frac{h(2.05) - h(2)}{2.05 - 2} = \frac{29.40875 - 30.4}{0.05} = \frac{-0.99125}{0.05} = -19.825
\]

#### Interval \([2, 2.001]\)

\[
a = 2, \, b = 2.001
\]

\[
h(2.001) = -4.9(2.001^2) + 50 = -4.9(4.004001) + 50 = -19.6196049 + 50 = 30.3803951
\]

\[
\text{AROC} = \frac{h(2.001) - h(2)}{2.001 - 2} = \frac{30.3803951 - 30.4}{0.001} = \frac{-0.0196049}{0.001} = -19.6049
\]

### Part (b): Estimating the Instantaneous Rate of Change

The instantaneous rate of change (IROC) at \( t = 2 \) is the limit of the average rate of change as the interval gets smaller and smaller, approaching 2 from both sides. The IROC can be estimated by observing the pattern in the AROCs:

- AROC on \([2, 2.5]\): -22.05
- AROC on \([2, 2.25]\): -20.825
- AROC on \([2, 2.05]\): -19.825
- AROC on \([2, 2.001]\): -19.6049

As the interval gets smaller, the AROC approaches -19.6. Thus, the IROC of the height with respect to time at \( t = 2 \) seconds is approximately:

\[
\text{IROC} \approx -19.6
\]

This is the derivative of \( h(t) = -4.9t^2 + 50 \) evaluated at \( t = 2 \).

### Conclusion

The IROC of the height at \( t = 2 \) seconds is -19.6 m/s, indicating that the ball is falling at a rate of 19.6 meters per second downward at that moment.

---

Would you like more details or have any questions?

### Additional Questions:
1. How would the IROC change if the initial height were different?
2. What is the physical meaning of the negative sign in the IROC?
3. How do we find the derivative of a quadratic function like this one?
4. What happens to the ball’s speed as time increases?
5. How would air resistance affect the IROC in real-world scenarios?

### Tip:
When estimating an IROC, the smaller the interval, the closer the AROC gets to the true derivative at a point.

The height of a ball dropped from a platform 50 meters off the ground, with initial velocity 0, at time t seconds, is given by: h(t) = −4.9t^2 + 50. a. Calculate the average rate of change of the height with respect to time on the intervals: [2, 2.5], [2, 2.25], [2, 2.05], [2, 2.001]. b. Based on part (a), what do you suspect is the IROC of the height with respect to time at t = 2?

Learn how to calculate the average and instantaneous rate of change for the height of a ball dropped from 50 meters, using the function h(t) = −4.9t^2 + 50. Includes step-by-step calculations for multiple intervals and an estimation of the IROC at t = 2. Suitable for high school students in Grades 10-12 studying quadratic functions and kinematics.

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