Question

NASA launches a rockets at t=0 seconds. Its height, in meters above sea level, as a function of time is given by h(t)=-4.9t squared +49t+283. 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.

Answer 1

To determine the time of splashdown and the maximum height of the rocket, we start with the height function given by:

\[
h(t) = -4.9t^2 + 49t + 283
\]

### Part 1: Time of Splashdown
The splashdown occurs when the height \( h(t) \) is equal to 0. We need to solve the equation:

\[
-4.9t^2 + 49t + 283 = 0
\]

We can use the quadratic formula to find the values of \( t \):

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = -4.9 \), \( b = 49 \), and \( c = 283 \). Now we calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = 49^2 - 4 \cdot (-4.9) \cdot 283
\]

Calculating \( 49^2 \):

\[
49^2 = 2401
\]

Now, calculate \( -4 \cdot (-4.9) \cdot 283 \):

\[
-4 \cdot (-4.9) \cdot 283 = 19.6 \cdot 283 = 5554.8
\]

So,

\[
b^2 - 4ac = 2401 + 5554.8 = 7955.8
\]

Now substituting back into the quadratic formula:

\[
t = \frac{-49 \pm \sqrt{7955.8}}{2 \cdot -4.9}
\]

Calculating \( \sqrt{7955.8} \):

\[
\sqrt{7955.8} \approx 89.18
\]

Now substituting:

\[
t = \frac{-49 \pm 89.18}{-9.8}
\]

Calculating the two possible solutions for \( t \):

1. \( t_1 = \frac{-49 + 89.18}{-9.8} \)
2. \( t_2 = \frac{-49 - 89.18}{-9.8} \)

Calculating \( t_1 \):

\[
t_1 = \frac{40.18}{-9.8} \approx -4.1 \quad (\text{Not a valid solution as time cannot be negative})
\]

Calculating \( t_2 \):

\[
t_2 = \frac{-138.18}{-9.8} \approx 14.1
\]

Thus, the splashdown occurs at approximately:

\[
\text{The rocket splashes down after } 14.1 \text{ seconds.}
\]

### Part 2: Maximum Height
The maximum height of a parabolic function \( h(t) = at^2 + bt + c \) occurs at the vertex. The time at which the maximum height occurs can be found using:

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

Substituting \( a = -4.9 \) and \( b = 49 \):

\[
t = -\frac{49}{2 \cdot -4.9} = \frac{49}{9.8} \approx 5
\]

Now to find the maximum height, we substitute \( t = 5 \) back into the height function:

\[
h(5) = -4.9(5^2) + 49(5) + 283
\]

Calculating \( 5^2 = 25 \):

\[
h(5) = -4.9(25) + 245 + 283
\]

Now calculating:

\[
h(5) = -122.5 + 245 + 283 = 405.5
\]

So, the maximum height the rocket reaches is:

\[
\text{The rocket peaks at } 405.5 \text{ meters above sea level.}
\]

### Summary of Answers
- The rocket splashes down after **14.1 seconds**.
- The rocket peaks at **405.5 meters** above sea level.