Question

nasa launches a rocket at $t = 0$ seconds. its height, in meters above sea - level, as a function of time is given by $h(t)=-4.9t^{2}+103t + 206$. 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.

Explanation:

Step1: Find splash - down time

Set $h(t)=0$, so we have the quadratic equation $-4.9t^{2}+103t + 206 = 0$. Use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, where $a=-4.9$, $b = 103$, and $c = 206$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(103)^{2}-4\times(-4.9)\times206=103^{2}+4\times4.9\times206=10609+4037.6 = 14646.6$.
Then $t=\frac{-103\pm\sqrt{14646.6}}{-9.8}=\frac{-103\pm121}{-9.8}$.
We get two solutions for $t$: $t_1=\frac{-103 + 121}{-9.8}=\frac{18}{-9.8}\approx - 1.84$ (rejected since time cannot be negative) and $t_2=\frac{-103-121}{-9.8}=\frac{-224}{-9.8}\approx22.86$.

Step2: Find the peak height

The function $h(t)=-4.9t^{2}+103t + 206$ is a quadratic function. The $t$ - value of the vertex of a quadratic function $y = ax^{2}+bx + c$ is $t=-\frac{b}{2a}$.
For $h(t)$, $a=-4.9$ and $b = 103$, so $t=-\frac{103}{2\times(-4.9)}=\frac{103}{9.8}\approx10.51$.
Substitute $t = \frac{103}{9.8}$ into $h(t)$:
$h(\frac{103}{9.8})=-4.9\times(\frac{103}{9.8})^{2}+103\times\frac{103}{9.8}+206$.
$h(\frac{103}{9.8})=-4.9\times\frac{103^{2}}{9.8^{2}}+\frac{103^{2}}{9.8}+206$.
$h(\frac{103}{9.8})=\frac{-4.9\times103^{2}+103^{2}\times9.8}{9.8^{2}}+206$.
$h(\frac{103}{9.8})=\frac{103^{2}( - 4.9+9.8)}{9.8^{2}}+206$.
$h(\frac{103}{9.8})=\frac{103^{2}\times4.9}{9.8^{2}}+206$.
$h(\frac{103}{9.8})=\frac{10609\times4.9}{96.04}+206$.
$h(\frac{103}{9.8}) = 540.25+206=746.25$.

Answer:

The rocket splashes down after approximately 22.86 seconds.
The rocket peaks at 746.25 meters above sea - level.