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.

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

\[ h(t) = -4.9t^2 + 265t + 241. \]

### Part 1: Time of Splashdown

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

\[ -4.9t^2 + 265t + 241 = 0. \]

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

1. Calculate the discriminant:

   \[
   b^2 - 4ac = 265^2 - 4 \times (-4.9) \times 241.
   \]

2. Find \( t \) 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:

\[
t = -\frac{b}{2a}.
\]

Substituting \( a \) and \( b \):

1. Calculate the time of peak height:

   \[
   t = -\frac{265}{2 \times -4.9}.
   \]

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

### Calculations

1. **Discriminant:**

   \[
   b^2 = 265^2 = 70225,
   \]
   \[
   4ac = 4 \times 4.9 \times 241 \approx 4717.6.
   \]
   \[
   \text{Discriminant} = 70225 + 4717.6 \approx 74942.6.
   \]

2. **Quadratic Formula:**

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

3. **Vertex for Max Height:**

   \[
   t = -\frac{265}{2 \times -4.9} \approx 26.98 \text{ seconds}.
   \]

4. **Max Height:**

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

### 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.

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

Explore the motion of a rocket launched by NASA, where its height is modeled by a quadratic equation. Learn to calculate the splashdown time of approximately 58.47 seconds and the peak height of approximately 3995.09 meters above sea-level. This content is ideal for high school students studying algebra and physics.

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