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}+115t + 304. assuming that the rocket will splash down into the ocean, at what time does splashdown occur? round the answer to the nearest tenth. the rocket splashes down after seconds. how high above sea - level does the rocket get at its peak? round the answer to the nearest tenth. 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 $-4.9t^{2}+115t + 304 = 0$. Use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, where $a=-4.9$, $b = 115$, and $c = 304$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(115)^{2}-4\times(-4.9)\times304=13225+5958.4 = 19183.4$.
Then $t=\frac{-115\pm\sqrt{19183.4}}{-9.8}=\frac{-115\pm138.5}{-9.8}$.
We get two solutions for $t$: $t_1=\frac{-115 + 138.5}{-9.8}=\frac{23.5}{-9.8}\approx - 2.4$ and $t_2=\frac{-115-138.5}{-9.8}=\frac{-253.5}{-9.8}\approx25.9$. Since time cannot be negative, the splash - down time is $t\approx25.9$ seconds.

Step2: Find the peak height

The function $h(t)=-4.9t^{2}+115t + 304$ 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 $b = 115$, so $t=-\frac{115}{2\times(-4.9)}=\frac{115}{9.8}\approx11.73$.
Substitute $t = 11.73$ into the height function $h(t)$: $h(11.73)=-4.9\times(11.73)^{2}+115\times11.73+304=-4.9\times137.59+1358+304=-674.2+1358+304 = 987.8$.

Answer:

The rocket splashes down after 25.9 seconds.
The rocket peaks at 987.8 meters above sea - level.