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The problem asks for the average rate at which the object falls during the first 3 seconds, using the given height function \( h(t) = 300 - 16t^2 \). 

### To find the average rate of change, we can use the formula:
\[
\text{Average rate of change} = \frac{h(t_2) - h(t_1)}{t_2 - t_1}
\]
Where:
- \( t_1 = 0 \) (the initial time)
- \( t_2 = 3 \) (after 3 seconds)

So, the average rate of change for the height during the first 3 seconds is:
\[
\frac{h(3) - h(0)}{3 - 0}
\]
This expression corresponds to the option:
\[
\frac{h(3) - h(0)}{3}
\]
which is the last choice in the list (the fourth option).

Would you like a detailed breakdown of the calculations, or do you have any other questions? Here are some related questions to consider:

1. What does the average rate of change represent physically in this context?
2. How do you calculate \( h(3) \) and \( h(0) \) from the given function?
3. How is the average rate of change different from the instantaneous rate of change?
4. What would happen if the object was dropped from a different height?
5. How can you interpret the negative sign in the rate of change?

**Tip**: The average rate of change over an interval gives the object's overall speed, while the instantaneous rate would tell you its exact speed at a given moment.

The height, h, of a falling object t seconds after it is dropped from a platform 300 feet above the ground is modeled by the function h(t) = 300 - 16t^2. Which expression could be used to determine the average rate at which the object falls during the first 3 seconds of its fall?

Determine the average rate at which an object falls during the first 3 seconds using the height function h(t) = 300 - 16t^2. This problem involves quadratic functions and average rate of change, suitable for students in Grades 9-12.

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