The given problem asks to find the average velocity of a rocket launched vertically upwards over a time interval $[1, 1 + h]$, where $h > 0$.

We are given the equation for the position of the rocket at time $t$ as: $s(t) = -16t^2 + 100t$ This is a quadratic function representing the rocket's position $s(t)$ in feet, where $s = 0$ corresponds to the ground.

Average Velocity Formula:

The average velocity of the rocket over the interval $[1, 1 + h]$ is given by: $\text{Average Velocity} = \frac{s(1 + h) - s(1)}{(1 + h) - 1} = \frac{s(1 + h) - s(1)}{h}$ Now, let’s compute $s(1 + h)$ and $s(1)$.

1. Compute $s(1)$: $s(1) = -16(1)^2 + 100(1) = -16 + 100 = 84$

2. Compute $s(1 + h)$: Substituting $t = 1 + h$ into the position function: $s(1 + h) = -16(1 + h)^2 + 100(1 + h)$ Expanding both terms: $s(1 + h) = -16(1 + 2h + h^2) + 100(1 + h)$ $s(1 + h) = -16(1 + 2h + h^2) + 100 + 100h$ $s(1 + h) = -16 - 32h - 16h^2 + 100 + 100h$ Simplifying: $s(1 + h) = 84 + 68h - 16h^2$

3. Average velocity: Now, apply these values to the average velocity formula: $\text{Average Velocity} = \frac{s(1 + h) - s(1)}{h} = \frac{(84 + 68h - 16h^2) - 84}{h}$ Simplifying: $\text{Average Velocity} = \frac{68h - 16h^2}{h}$ $\text{Average Velocity} = 68 - 16h$

Final Answer:

The average velocity of the rocket over the time interval $[1, 1 + h]$ is: $\boxed{68 - 16h} \, \text{ft/s}.$

Would you like more details or clarifications on any part of this solution?

Here are 5 questions to extend this topic:

1. How does the acceleration due to gravity affect the equation of motion?
2. How would air resistance change the result for average velocity?
3. What would the velocity of the rocket be at $t = 1$?
4. How can you determine the maximum height reached by the rocket?
5. What is the significance of the coefficient $-16$ in the equation for position?

Tip: The derivative of the position function gives the instantaneous velocity, which can help when analyzing motion at specific moments in time.