Math Problem Statement

The function h(t) = -4.87t^2 + 18.75t is used to model the height of an object projected in the air. What are the domain and range of the function h(t)? Round values to the nearest hundredth.

Solution

The problem shows a quadratic function representing the height of an object in the air:

h(t)=4.87t2+18.75th(t) = -4.87t^2 + 18.75t

where h(t)h(t) is the height in meters, and tt is the time in seconds. The task is to find the domain and range of this function, rounding values to the nearest hundredth.

1. Domain:

The domain is the set of time values (tt) for which the height is defined. Since the object is projected into the air, the object will start at t=0t = 0 and will land back on the ground when h(t)=0h(t) = 0.

  • To find when the object hits the ground, set h(t)=0h(t) = 0 and solve for tt:

0=4.87t2+18.75t0 = -4.87t^2 + 18.75t

Factor the quadratic equation:

0=t(4.87t+18.75)0 = t(-4.87t + 18.75)

This gives two solutions:

  • t=0t = 0 (when the object is launched),
  • t=18.754.873.85t = \frac{18.75}{4.87} \approx 3.85 seconds (when the object hits the ground).

Thus, the domain is [0,3.85][0, 3.85].

2. Range:

The range is the set of height values (hh) that the object reaches. The maximum height occurs at the vertex of the parabola. To find the time at which the maximum height occurs, use the formula for the vertex:

tvertex=b2a=18.752(4.87)1.92t_{\text{vertex}} = \frac{-b}{2a} = \frac{-18.75}{2(-4.87)} \approx 1.92

Substitute t=1.92t = 1.92 into the function h(t)h(t) to find the maximum height:

h(1.92)=4.87(1.92)2+18.75(1.92)18.05metersh(1.92) = -4.87(1.92)^2 + 18.75(1.92) \approx 18.05 \, \text{meters}

Thus, the range is [0,18.05][0, 18.05], since the height starts at 0 meters and reaches a maximum of 18.05 meters.

Final Answer:

  • Domain: [0,3.85][0, 3.85] seconds
  • Range: [0,18.05][0, 18.05] meters

Let me know if you'd like more details or explanations.


Questions to expand:

  1. How is the vertex of a quadratic function determined?
  2. Why is the height 0 when the object hits the ground?
  3. What does the coefficient of t2t^2 tell us about the direction of the parabola?
  4. How would the problem change if the initial velocity (coefficient of tt) were different?
  5. Can you apply a similar method to model other types of motion (e.g., objects dropped from a height)?

Tip: In quadratic equations modeling physical scenarios, the vertex often represents a maximum or minimum in the context of height or distance.

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Math Problem Analysis

Mathematical Concepts

Quadratic Function
Domain and Range
Projectile Motion

Formulas

h(t) = -4.87t^2 + 18.75t
Vertex Formula: t = -b / (2a)
Quadratic Zero: t = 0 and t = 3.85

Theorems

Quadratic Function Vertex
Projectile Motion Equation

Suitable Grade Level

Grades 9-12