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) = – 4.9t2 + 235t + 318. Assuming that the backet 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.

Answer 1

The height of the equation is the sea-level hight, so, whenit splashes into the ocean, h(t) = 0. Let's wirte that down:

`\begin{gathered} h(t)=0 \\ 0=-4.9t^2+235t+318 \end{gathered}`

So, the find t for when the rocket splashes into the ocean, we just have to solve for the roots.

We can use Bhaskara's Formula for that:

`\begin{gathered} a=-4.9 \\ b=235 \\ c=318 \\ t=\frac{-235\pm\sqrt[]{235^2-4\cdot(-4.9)\cdot318}}{2\cdot(-4.9)} \end{gathered}`

Let's calculate, then:

`\begin{gathered} t=\frac{-235\pm\sqrt[]{235^2-4\cdot(-4.9)\cdot318}}{2\cdot(-4.9)}=\frac{-235\pm\sqrt[]{55225+6232.8}}{-9.8} \\ t=\frac{-235\pm\sqrt[]{61457.8}}{-9.8}\approx(-235\pm247.9068)/(-9.8) \\ t_1=(-235+247.9068)/(-9.8)=(12.9068)/(-9.8)\approx-1.3170 \\ t_2=(-235-247.9068)/(-9.8)=(-482.9068)/(-9.8)\approx49.276 \end{gathered}`

Since the launch was at t = 0, we can't have a negative value of t, so our answer is t = 49.276 s.

To calculate how high above the sea-level the rocket gets, we can firts find the x value of the vertex of the equation, which will be the highest point. This will be our t.

`t_(\max )=x_v=-(b)/(2a)=-(235)/(2\cdot(-4.9))=-(235)/(-9.8)\approx23.9796`

Now, we input this t into the equation for the height:

`\begin{gathered} h(t_(\max ))=-4.9t^2_(\max )+235t_(\max )+318=-4.9\cdot(23.9796)^2+235\cdot23.9796+318 \\ h(t_(\max ))=-4.9\cdot575.0212+5635.206+318=5953.206-2817.60388=3135.602 \end{gathered}`

So, the rocket peacks at 3135.602 meters above the sea-level.