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}+67t + 157. 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. question help: video 1 video 2

Explanation:

Step1: Find splash - down time

Set $h(t)=0$, so we have the quadratic equation $-4.9t^{2}+67t + 157 = 0$. The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=-4.9$, $b = 67$, and $c = 157$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(67)^{2}-4\times(-4.9)\times157=4489+3077.2 = 7566.2$.
Then, $t=\frac{-67\pm\sqrt{7566.2}}{2\times(-4.9)}=\frac{-67\pm86.99}{-9.8}$.
We get two solutions for $t$: $t_1=\frac{-67 + 86.99}{-9.8}=\frac{19.99}{-9.8}\approx - 2.04$ and $t_2=\frac{-67-86.99}{-9.8}=\frac{-153.99}{-9.8}\approx15.71$. Since time cannot be negative, the splash - down time is $t\approx15.71$ seconds.

Step2: Find the peak height

The function $h(t)=-4.9t^{2}+67t + 157$ 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 $h(t)$, $t=-\frac{67}{2\times(-4.9)}=\frac{67}{9.8}\approx6.84$ seconds.
Substitute $t = \frac{67}{9.8}$ into $h(t)$:
$h(\frac{67}{9.8})=-4.9\times(\frac{67}{9.8})^{2}+67\times\frac{67}{9.8}+157$.
$h(\frac{67}{9.8})=-4.9\times\frac{4489}{96.04}+ \frac{4489}{9.8}+157$.
$h(\frac{67}{9.8})=-\frac{22096.1}{96.04}+\frac{4489}{9.8}+157$.
$h(\frac{67}{9.8})=-230.1+458.06+157$.
$h(\frac{67}{9.8})\approx384.96$ meters.

Answer:

The rocket splashes down after approximately 15.71 seconds.
The rocket peaks at approximately 384.96 meters above sea - level.