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 25.9 seconds. how high above sea - level does the rocket get at its peak? round the answer to the nearest tenth. the rocket peaks at 987.8 meters above sea - level.

Explanation:

Step1: Identify the formula for vertex of parabola

For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate of the vertex is $t=-\frac{b}{2a}$. Here, $a=-4.9$, $b = 115$, $c = 304$ in $h(t)=-4.9t^{2}+115t + 304$.

Step2: Calculate the time at which the rocket reaches its peak

$t=-\frac{115}{2\times(-4.9)}=\frac{115}{9.8}\approx11.73$ seconds.

Step3: Calculate the height at the peak

Substitute $t = \frac{115}{9.8}$ into $h(t)=-4.9t^{2}+115t + 304$.
$h(\frac{115}{9.8})=-4.9\times(\frac{115}{9.8})^{2}+115\times\frac{115}{9.8}+304$
$=-4.9\times\frac{115^{2}}{9.8^{2}}+\frac{115^{2}}{9.8}+304$
$=-\frac{115^{2}}{19.6}+\frac{115^{2}}{9.8}+304$
$=\frac{-115^{2}+2\times115^{2}}{19.6}+304$
$=\frac{115^{2}}{19.6}+304$
$=\frac{13225}{19.6}+304$
$=674.74 + 304$
$=978.74\approx978.7$ meters.

Answer:

The rocket peaks at 978.7 meters above sea - level.