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.

Explanation:

Step1: Find splash - down time

The rocket splashes down when $h(t)=0$. So we set up the quadratic equation $-4.9t^{2}+67t + 157 = 0$. For a quadratic equation $ax^{2}+bx + c = 0$ ($a=-4.9$, $b = 67$, $c = 157$), the quadratic formula is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
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 $t=-\frac{b}{2a}$. Here, $a=-4.9$ and $b = 67$, so $t=-\frac{67}{2\times(-4.9)}=\frac{67}{9.8}\approx6.84$ seconds.
Substitute $t = 6.84$ into the height function $h(t)$: $h(6.84)=-4.9\times(6.84)^{2}+67\times6.84 + 157=-4.9\times46.79+468.28+157=-229.27+468.28+157=396.01$ meters.

Answer:

The rocket splashes down after approximately 15.71 seconds.
The rocket peaks at approximately 396.01 meters above sea - level.