Question

a model rocket is launched from a platform. its height, h metres, as a function of the elapsed time, t seconds, is given by the function h(t)= - 4.9t²+180t + 2. use a graphing calculator to determine the following: a) the maximum height of the rocket vertex=(18.36735,1655.0612) maximum = 1655.06 m b) the time required for the rocket to reach its maximum height maximum when x = 18.36735≈18.37 s c) the time required for the rocket to reach the ground y = 0 when x = 36.745802≈36.75 s

Explanation:

Step1: Recall vertex - formula for a quadratic function

For a quadratic function $h(t)=at^{2}+bt + c$ (in this case $a=-4.9$, $b = 180$, $c = 2$), the $t$ - value of the vertex (time at maximum height) is given by $t=-\frac{b}{2a}$.
$t=-\frac{180}{2\times(-4.9)}=\frac{180}{9.8}\approx18.37$ s.

Step2: Find the maximum height

Substitute $t = \frac{180}{9.8}$ into the function $h(t)=-4.9t^{2}+180t + 2$.
$h(\frac{180}{9.8})=-4.9\times(\frac{180}{9.8})^{2}+180\times\frac{180}{9.8}+2$
$=-4.9\times\frac{180^{2}}{9.8^{2}}+\frac{180^{2}}{9.8}+2$
$=-\frac{180^{2}}{19.6}+\frac{180^{2}}{9.8}+2$
$=\frac{180^{2}}{19.6}+2=\frac{32400}{19.6}+2\approx1655.06$ m.

Step3: Find the time when the rocket reaches the ground

Set $h(t)=0$, so $-4.9t^{2}+180t + 2 = 0$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, where $a=-4.9$, $b = 180$, $c = 2$.
$t=\frac{-180\pm\sqrt{180^{2}-4\times(-4.9)\times2}}{2\times(-4.9)}=\frac{-180\pm\sqrt{32400 + 39.2}}{-9.8}=\frac{-180\pm\sqrt{32439.2}}{-9.8}$.
We take the positive root since time cannot be negative. $t\approx36.75$ s.

Answer:

a) The maximum height of the rocket is approximately $1655.06$ m.
b) The time required for the rocket to reach its maximum height is approximately $18.37$ s.
c) The time required for the rocket to reach the ground is approximately $36.75$ s.