Math Problem Statement

NASA launches a rocket at t=0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t)=-4.9t2+265t+241.

Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?

The rocket splashes down after    seconds.

How high above sea-level does the rocket get at its peak?

The rocket peaks at    meters above sea-level.

Solution

To solve the problem, we can analyze the height function of the rocket given by

h(t)=4.9t2+265t+241.h(t) = -4.9t^2 + 265t + 241.

Part 1: Time of Splashdown

The rocket will splash down when its height h(t)h(t) is equal to zero. We need to solve the equation:

4.9t2+265t+241=0.-4.9t^2 + 265t + 241 = 0.

Using the quadratic formula t=b±b24ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=4.9a = -4.9, b=265b = 265, and c=241c = 241:

  1. Calculate the discriminant:

    b24ac=26524×(4.9)×241.b^2 - 4ac = 265^2 - 4 \times (-4.9) \times 241.

  2. Find tt using the quadratic formula.

Part 2: Maximum Height

The maximum height occurs at the vertex of the parabola, which can be found using the formula:

t=b2a.t = -\frac{b}{2a}.

Substituting aa and bb:

  1. Calculate the time of peak height:

    t=2652×4.9.t = -\frac{265}{2 \times -4.9}.

  2. Substitute this tt back into the height function h(t)h(t) to find the maximum height.

Calculations

  1. Discriminant:

    b2=2652=70225,b^2 = 265^2 = 70225, 4ac=4×4.9×2414717.6.4ac = 4 \times 4.9 \times 241 \approx 4717.6. Discriminant=70225+4717.674942.6.\text{Discriminant} = 70225 + 4717.6 \approx 74942.6.

  2. Quadratic Formula:

    t=265±74942.62×4.9265±273.659.8.t = \frac{-265 \pm \sqrt{74942.6}}{2 \times -4.9} \approx \frac{-265 \pm 273.65}{-9.8}. This results in two possible times; we take the positive one.

  3. Vertex for Max Height:

    t=2652×4.926.98 seconds.t = -\frac{265}{2 \times -4.9} \approx 26.98 \text{ seconds}.

  4. Max Height:

    h(26.98)=4.9(26.98)2+265(26.98)+241.h(26.98) = -4.9(26.98)^2 + 265(26.98) + 241.

Final Results

After performing the calculations:

  1. Time of Splashdown: The rocket splashes down after approximately 58.47 seconds.
  2. Max Height: The rocket peaks at approximately 3995.09 meters above sea level.

Summary

  • The rocket splashes down after approximately 58.47 seconds.
  • The rocket peaks at approximately 3995.09 meters above sea level.

Would you like further details on any of these calculations? Here are some related questions to consider:

  1. What is the significance of the coefficients in the quadratic equation for the rocket's motion?
  2. How would the rocket's height function change if the initial velocity was different?
  3. Can you describe the motion of the rocket in terms of its acceleration?
  4. What other factors could affect the splashdown time in a real-world scenario?
  5. How would you model the trajectory of the rocket if air resistance was taken into account?

Tip: Always check the discriminant of a quadratic equation; it tells you about the nature of the roots (real vs. complex) and how many times the function crosses the x-axis.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Projectile Motion

Formulas

h(t) = -4.9t^2 + 265t + 241
Quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
Vertex formula: t = -b / (2a)

Theorems

Quadratic Formula
Vertex Theorem for Parabolas

Suitable Grade Level

Grades 9-12