Math Problem Statement

The height in inches of a frog's jump is modeled by the equation h(t) = 60t -75t^2 where the time t m afterit jumped is measured in seconds. a) Find h(0) and h(0.8) .What does these values mean in terms of the frog's jump b) Determine the time it takes for the frog to reach its maximum height. What was the maximum height. c) Determine how long the frog is in the air. d) Determine the height that frog jumped from . d) What domain could be used to descrie the frog's jump? Range ?

Solution

Let's analyze the problem step by step.

a) Finding h(0)h(0) and h(0.8)h(0.8)

The height of the frog's jump is given by the equation: h(t)=60t75t2h(t) = 60t - 75t^2

  1. Finding h(0)h(0): h(0)=60(0)75(02)=0h(0) = 60(0) - 75(0^2) = 0 This means that at t=0t = 0 seconds (the moment it jumps), the frog is at ground level (height = 0 inches).

  2. Finding h(0.8)h(0.8): h(0.8)=60(0.8)75(0.82)=4875(0.64)=4848=0h(0.8) = 60(0.8) - 75(0.8^2) = 48 - 75(0.64) = 48 - 48 = 0 This means that at t=0.8t = 0.8 seconds, the frog returns to the ground level after reaching its maximum height.

b) Maximum Height and Time to Reach It

To find the time it takes to reach maximum height, we can use the vertex formula for a quadratic equation, t=b2at = -\frac{b}{2a}.

Here, a=75a = -75 and b=60b = 60: t=602(75)=60150=0.4 secondst = -\frac{60}{2(-75)} = \frac{60}{150} = 0.4 \text{ seconds}

Maximum Height: Now, plug t=0.4t = 0.4 back into h(t)h(t): h(0.4)=60(0.4)75(0.42)=2475(0.16)=2412=12 inchesh(0.4) = 60(0.4) - 75(0.4^2) = 24 - 75(0.16) = 24 - 12 = 12 \text{ inches} Thus, the maximum height is 12 inches.

c) Time in the Air

The frog is in the air until it returns to the ground, which we determined is at h(t)=0h(t) = 0. We already found that h(0)=0h(0) = 0 and h(0.8)=0h(0.8) = 0. Therefore, the total time the frog is in the air is 0.8 seconds.

d) Height the Frog Jumped From

The height from which the frog jumped is h(0)=0h(0) = 0 inches, meaning it jumped from ground level.

e) Domain and Range of the Frog's Jump

  • Domain: The time tt must be non-negative and should not exceed the time it takes to hit the ground. Thus, the domain can be expressed as: t[0,0.8]t \in [0, 0.8]

  • Range: The height h(t)h(t) varies from 00 (at t=0t = 0 and t=0.8t = 0.8) to the maximum height of 1212 inches. Thus, the range is: h(t)[0,12]h(t) \in [0, 12]

Summary of Results

  1. h(0)=0h(0) = 0 (frog is at ground level at the jump)
  2. h(0.8)=0h(0.8) = 0 (frog returns to ground level)
  3. Time to reach maximum height: 0.4 seconds, Maximum height: 12 inches
  4. Time in the air: 0.8 seconds
  5. Jumped from: 0 inches (ground level)
  6. Domain: [0,0.8][0, 0.8], Range: [0,12][0, 12]

Do you want further details on any part, or do you have any questions? Here are some related questions to consider:

  1. What is the significance of the coefficients in the equation h(t)=60t75t2h(t) = 60t - 75t^2?
  2. How would the graph of this function look like?
  3. What would happen if the coefficients were changed?
  4. Can you find the time taken for the frog to reach half its maximum height?
  5. How would you describe the motion of the frog in terms of acceleration?

Tip: When dealing with quadratic functions, the vertex form can often help identify maximum or minimum values quickly.

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Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Projectile Motion

Formulas

h(t) = 60t - 75t^2
t = -b/(2a) for maximum height

Theorems

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Suitable Grade Level

Grades 9-11