Let's analyze the problem step-by-step to determine the splashdown time and the peak height of the rocket.
Given the height function of the rocket:
[tex]\[ h(t) = -4.9 t^2 + 172 t + 388 \][/tex]
### 1. Time of Splashdown
Splashdown occurs when the rocket's height [tex]\( h(t) \)[/tex] reaches zero. We need to find the value of [tex]\( t \)[/tex] when [tex]\( h(t) = 0 \)[/tex].
[tex]\[ -4.9 t^2 + 172 t + 388 = 0 \][/tex]
This is a quadratic equation in standard form [tex]\( at^2 + bt + c = 0 \)[/tex]. The roots of this equation can be found using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = -4.9 \)[/tex], [tex]\( b = 172 \)[/tex], and [tex]\( c = 388 \)[/tex].
Upon solving this quadratic equation (including consideration for the positive root):
[tex]\[ t_{\text{total}} = 37.23 \][/tex]
So, the rocket splashes down after [tex]\( \boxed{37.23} \)[/tex] seconds.
### 2. Peak Height
The peak height of the rocket occurs at the vertex of the parabola described by the quadratic function. The time [tex]\( t_{\text{peak}} \)[/tex] at which the peak occurs can be found using the formula for the vertex of a parabola:
[tex]\[ t_{\text{peak}} = -\frac{b}{2a} \][/tex]
Substituting [tex]\( a = -4.9 \)[/tex] and [tex]\( b = 172 \)[/tex]:
[tex]\[ t_{\text{peak}} = -\frac{172}{2 \cdot (-4.9)} = 17.55 \][/tex]
Now, we need to find the height at this time. This is done by substituting [tex]\( t = 17.55 \)[/tex] back into the height function:
[tex]\[ h(17.55) = -4.9 (17.55)^2 + 172(17.55) + 388 \][/tex]
Solving this, we get the peak height:
[tex]\[ \text{peak height} = 1897.39 \][/tex]
Thus, the rocket peaks at [tex]\( \boxed{1897.39} \)[/tex] meters above sea level.
Therefore:
- The rocket splashes down after [tex]\( \boxed{37.23} \)[/tex] seconds.
- The rocket peaks at [tex]\( \boxed{1897.39} \)[/tex] meters above sea level.
