Question

Assuming that the rocket will splash down into the ocean, at what time does splash down occur?

Answer 1

The height in meters is given as a function of t (time) as;

`h(t)=-4.9t^2+268t+416`

First, let's solve the quadratic function using quadratic formula given as;

`\begin{gathered} t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where a=-4.9} \\ b=268\text{ and c=416} \end{gathered}`
`\begin{gathered} t=\frac{-268\pm\sqrt[]{268^2-4(-4.9)(416)}}{2(-4.9)} \\ t=\frac{-268\pm\sqrt[]{71824+8153.6}}{-9.8} \\ t=(-268\pm282.80)/(-9.8) \\ t=(-268+282.80)/(-9.8)\text{ or t=}(-268-282.80)/(-9.8) \\ t=-1.51\text{ or 56.20} \end{gathered}`

Since a time cannot be a negative value. Hence, the rocket splashes down after 56.20seconds.

Also, to find the peak, let's find the time

`\begin{gathered} h(t)=-4.9t^2+268t+416 \\ h^(\prime)(t)=-9.8t+268 \\ At\text{ the peak, h'(t)=0} \\ \text{That is;} \\ -9.8t+268=0 \\ t=(268)/(9.8) \\ t=27.35\text{seconds} \end{gathered}`

Thus the h(t) above the sea level is at time t=27.35seconds is;

`\begin{gathered} h(27.35)=-4.9(27.35)^2+268(27.35)+416 \\ h(27.35)=4080.49m \end{gathered}`

The rocket peaks at 4080.49meters above the sea level