Question

question 18
the height of a rocket that is launched is modeled by the function h(t)= - 4.9t^2 + 1500t + 420 where the height is meters above sea - level, t seconds after launch.
a) (2 pts) what is the maximum height the rocket reaches? (round your answer to two decimal places if necessary.)
select an answer
b) (2 pts) at what time does the rocket reach its maximum height? (round your answer to two decimal places if necessary.)
select an answer
c) (2 pts) what is the height of the rocket at the time of the launch? (round your answer to two decimal places if necessary.)
select an answer
question help: video post to forum
submit question

Explanation:

Step1: Identify the function type

The height function $h(t)=-4.9t^{2}+1500t + 420$ is a quadratic function in the form $y = ax^{2}+bx + c$ where $a=-4.9$, $b = 1500$, $c = 420$.

Step2: Find the time of maximum height

For a quadratic function $y=ax^{2}+bx + c$, the $x$-coordinate (in our case $t$ - coordinate) of the vertex (where maximum or minimum occurs) is given by $t=-\frac{b}{2a}$.
$t=-\frac{1500}{2\times(-4.9)}=\frac{1500}{9.8}\approx153.06$ seconds.

Step3: Find the maximum height

Substitute $t = \frac{1500}{9.8}$ into the height - function $h(t)$.
$h(\frac{1500}{9.8})=-4.9\times(\frac{1500}{9.8})^{2}+1500\times\frac{1500}{9.8}+420$
$=-4.9\times\frac{1500^{2}}{9.8^{2}}+\frac{1500^{2}}{9.8}+420$
$=\frac{-4.9\times1500^{2}+1500^{2}\times9.8}{9.8^{2}}+420$
$=\frac{1500^{2}(9.8 - 4.9)}{9.8^{2}}+420$
$=\frac{1500^{2}\times4.9}{9.8^{2}}+420$
$=\frac{1500^{2}}{19.6}+420$
$=\frac{2250000}{19.6}+420\approx114795.92 + 420=115215.92$ meters.

Step4: Find the height at launch

At the time of launch $t = 0$. Substitute $t = 0$ into $h(t)$.
$h(0)=-4.9\times0^{2}+1500\times0 + 420=420.00$ meters.

Answer:

a) $115215.92$ meters
b) $153.06$ seconds
c) $420.00$ meters