NASA launches a rocket at [tex]t=0[/tex] seconds. Its height, in meters above sea level, as a function of time is given by [tex]h(t)=-4.9 t^2+367 t+276[/tex].

1. Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
- The rocket splashes down after [tex]\square[/tex] seconds.

2. How high above sea level does the rocket get at its peak?
- The rocket peaks at [tex]\square[/tex] meters above sea level.

Question Help: [tex]\square[/tex] Video 1 [tex]\square[/tex] Video 2



Answer :

To solve this problem, we need to find two key pieces of information from the given quadratic function [tex]\( h(t) = -4.9t^2 + 367t + 276 \)[/tex]:

1. The time of splashdown, which occurs when the height [tex]\( h(t) \)[/tex] is zero.
2. The maximum height achieved by the rocket, which is the peak of the quadratic function.

### Finding the Time of Splashdown

To find the time of splashdown, we solve the equation [tex]\( h(t) = 0 \)[/tex]:

[tex]\[ -4.9t^2 + 367t + 276 = 0 \][/tex]

This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -4.9 \)[/tex], [tex]\( b = 367 \)[/tex], and [tex]\( c = 276 \)[/tex].

Solving for [tex]\( t \)[/tex] using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:

1. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 367^2 - 4(-4.9)(276) \][/tex]

2. Calculate the roots:
[tex]\[ t = \frac{-367 \pm \sqrt{\Delta}}{2(-4.9)} \][/tex]

You will get two potential solutions for [tex]\( t \)[/tex]. However, the negative solution is not physically meaningful in this context, so we take the positive solution. After solving, we find:

The rocket splashes down after 75.6426 seconds.

### Finding the Peak Height

The peak height of the rocket is found at the vertex of the parabola represented by the quadratic function. The time at which this occurs can be found using the vertex formula for a parabola [tex]\( t = -\frac{b}{2a} \)[/tex]:

1. Compute the time to reach the peak height:
[tex]\[ t = -\frac{367}{2(-4.9)} \][/tex]
[tex]\[ t = \frac{367}{9.8} \][/tex]
[tex]\[ t \approx 37.449 \, \text{seconds} \][/tex]

2. Substitute this time back into the height function to get the peak height:
[tex]\[ h(t) = -4.9(37.449)^2 + 367(37.449) + 276 \][/tex]

Calculating this gives us the peak height:

The rocket peaks at 7147.888 meters above sea level.

Thus, the precise answers are:

- The rocket splashes down after 75.6426 seconds.
- The rocket peaks at 7147.888 meters above sea level.