To solve this, we use the laws of kinematics under constant acceleration due to gravity. Here's the step-by-step process:

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Given:

1. Total time to fall (t) = 5.0 s
2. Acceleration due to gravity (g) = 9.8 m/s²
3. At the maximum height, the final velocity (just before falling) is 0 m/s.

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Step 1: Relation Between Time and Height

The time for the rocket to fall from the maximum height can be modeled using the kinematic equation: $h = \frac{1}{2} g t^2$ Substitute the known values: $h = \frac{1}{2} (9.8) (5.0)^2$ $h = 0.5 \times 9.8 \times 25 = 122.5 \, \text{m}$

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Step 2: Approximation

The closest approximate height is 120 m, which corresponds to Option C.

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Final Answer:

C) 120 m

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Do you want me to explain further or address any related questions? 😊

Related Questions:

1. How do we derive the kinematic equation used here?
2. What happens if air resistance is considered?
3. How would the problem change if the total time was 6 seconds instead of 5?
4. What is the significance of maximum height in this problem?
5. How do you calculate the velocity of the rocket just before impact?

Tip: Always remember that the total fall time is crucial to deducing the maximum height of an object under gravity!