Math Problem Statement

The platform for the high dive is 39 feet above the water. A diver jumps from the platform and lands in the water after 1.1 seconds. The function H(s) represents the height of the diver after s seconds. Determine the practical domain and practical range of H(s) in this situation.

Solution

The problem you uploaded involves determining the practical domain and practical range for the function H(s)H(s), which represents the height of a diver above water after jumping from a platform 39 feet high.

Let's break this down:

  • Height function, H(s)H(s):
    • The platform is 39 feet high.
    • The diver lands in the water after 1.1 seconds. This means that at s=1.1s = 1.1 seconds, the height H(s)=0H(s) = 0.

Practical Domain:

  • The time ss starts from when the diver jumps, so the practical domain is from 0 to 1.1 seconds.
    • 0s1.10 \leq s \leq 1.1 seconds.

Practical Range:

  • The height of the diver starts at 39 feet and ends at 0 feet when the diver lands in the water.
    • 0H(s)390 \leq H(s) \leq 39 feet.

Final Answer:

  • Practical Domain: 0s1.10 \leq s \leq 1.1 seconds.
  • Practical Range: 0H(s)390 \leq H(s) \leq 39 feet.

Would you like more details or clarification on this?

Here are 5 related questions to consider:

  1. How do you find the practical domain for a general height function in physics problems?
  2. Why is it important to focus on the practical range rather than the theoretical range?
  3. How would the problem change if the diver jumped from a different height?
  4. How can we model H(s)H(s) more precisely with a function that includes gravity's effect?
  5. What other real-world scenarios involve similar domain and range considerations?

Tip: When determining the practical domain, always consider real-world constraints like time or distance.

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Math Problem Analysis

Mathematical Concepts

Functions
Domain and Range
Real-world Applications

Formulas

Height function H(s)

Theorems

Domain and Range in Real-World Contexts

Suitable Grade Level

Grades 8-10