To solve the problem, we need to calculate the average rate of change in height $H(t)$ over the given time intervals.

Formula for the Average Rate of Change:

$\text{Average Rate of Change} = \frac{\Delta H(t)}{\Delta t} = \frac{H(t_2) - H(t_1)}{t_2 - t_1}$

(a) From $t = 0$ to $t = 2.8$:

$\Delta H(t) = H(2.8) - H(0) = 84 - 0 = 84 \, \text{meters}$ $\Delta t = 2.8 - 0 = 2.8 \, \text{seconds}$ $\text{Average Rate of Change} = \frac{\Delta H(t)}{\Delta t} = \frac{84}{2.8} = 30 \, \text{meters per second}$

(b) From $t = 5.6$ to $t = 11.2$:

$\Delta H(t) = H(11.2) - H(5.6) = 0 - 140 = -140 \, \text{meters}$ $\Delta t = 11.2 - 5.6 = 5.6 \, \text{seconds}$ $\text{Average Rate of Change} = \frac{\Delta H(t)}{\Delta t} = \frac{-140}{5.6} = -25 \, \text{meters per second}$

Final Answers:

• (a) $30 \, \text{meters per second}$
• (b) $-25 \, \text{meters per second}$

Would you like a detailed explanation of how these values were derived, or do you have further questions?

Here are five related questions for further understanding:

1. How is the sign of the average rate of change interpreted in these calculations?
2. What does a negative average rate of change indicate about the rocket’s motion?
3. How does the average rate of change compare to the instantaneous rate of change?
4. Can you calculate the total distance traveled by the rocket during its entire flight?
5. How would the average rate of change change if the intervals were smaller?

Tip: The average rate of change is essentially the slope of the secant line between two points on a graph.