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) = -4.9t^2 + 361t + 108. Assuming that the rocket will splash down into the ocean, at what time does splashdown occur? How high above sea-level does the rocket get at its peak?

Solution

The image contains a quadratic function assignment that focuses on the motion of a rocket. The height of the rocket h(t)h(t) above sea level as a function of time tt is given by the equation:

h(t)=4.9t2+361t+108h(t) = -4.9t^2 + 361t + 108

There are two main parts to the question:

  1. Time of Splashdown: The rocket splashes into the ocean at t=73.9t = 73.9 seconds.
  2. Maximum Height: The rocket reaches its peak height of 6640.786640.78 meters.

Let me know if you'd like to explore how the peak height and time of splashdown were calculated, or if you'd like help with another specific part of the problem!

Here are 5 follow-up questions to think about:

  1. How is the formula for the height of the rocket derived?
  2. Why is there a negative coefficient for the t2t^2-term in the height equation?
  3. How can we calculate the time at which the rocket reaches its peak height?
  4. How would the equation change if the rocket's initial velocity or height were different?
  5. What would happen to the height-time graph if the gravity value (currently 4.9 m/s²) was different?

Tip:

When solving quadratic equations like this, the peak (or vertex) of the parabola can be found using the formula t=b2at = \frac{-b}{2a}, where the quadratic is in the form at2+bt+cat^2 + bt + c.

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Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Parabolas
Physics of Motion

Formulas

Quadratic function: h(t) = -4.9t^2 + 361t + 108
Vertex formula for maximum height: t = -b / 2a

Theorems

Quadratic Formula

Suitable Grade Level

Grades 9-11