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+121t+283 .Use the techniques of College Algebra (not graphing software) to demonstrate the following.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.

Answer 1

NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by

$h(t)=-4.9t^2+121t+283$

The sea level is represented by h = 0, therefore, to find the corresponding time when h splashes into the ocean we have to solve for t the following equation:

$h(t)=0\implies-4.9t^2+121t+283=0$

Using the quadratic formula, the solution for our problem is

$\begin{gathered} t=\frac{-(121)\pm\sqrt{(121)^2-4(-4.9)(283)}}{2(-4.9)} \\ =\frac{121\pm142.083778...}{9.8} \\ =\frac{121+142.083778...}{9.8} \\ =26.8452834694... \end{gathered}$

The rocket splashes after 26.845 seconds.

The maximum of this function happens at the root of the derivative. Differentiating our function, we have

$h^{\prime}(t)=-9.8t+121$

The root is

$-9.8t+121=0\implies t=\frac{121}{9.8}=12.347$

Then, the maximum height is

$h(12.347)=1029.99$

1029.99 meters above sea level.