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 v = 0 when x = 36.745802≈36.75 s

Explanation:

Step1: Identify the function

The height - time function of the rocket is $h(t)=-4.9t^{2}+180t + 2$. This is a quadratic function of the form $y = ax^{2}+bx + c$, where $a=-4.9$, $b = 180$, and $c = 2$.

Step2: Find the time to reach maximum height

For a quadratic function $y = ax^{2}+bx + c$, the $t$ - value of the vertex (time to reach maximum height) is given by $t=-\frac{b}{2a}$. Substituting $a=-4.9$ and $b = 180$ into the formula, we have $t=-\frac{180}{2\times(-4.9)}=\frac{180}{9.8}\approx18.37$ s.

Step3: Find the maximum height

Substitute $t = 18.37$ into the function $h(t)=-4.9t^{2}+180t + 2$.
$h(18.37)=-4.9\times(18.37)^{2}+180\times18.37 + 2$
$=-4.9\times337.46+3306.6+2$
$=-1653.554+3306.6 + 2$
$\approx1655.06$ m.

Step4: Find the time to reach the ground

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