Question

answer each question about pablo and maxs rocket and type the correct
coo ease remember to type in all caps with no spaces.

pablo and maxs rocket follows a
flight pattern approximately modeled
by the equation ( h ( t ) = - 8 t ^ { 2 } + 16 t + 2 )
where ( h ( t ) ) is their rockets height in
meters t seconds after launch.

1 how
many
seconds
was their rocket
in the air?

2 how many seconds
after launch did their
rocket reach its
maximum height?

3 what was their
rockets maximum
height, in meters?

how many
meters tall
was the
launch
platform?

round to hundredths
type the 4 - letter
code into the
answer box.
all caps, no spaces.

your answer

Explanation:

Step1: Find the time the rocket was in the air

Set \(h(t)=0\), so \(-8t^{2}+16t + 2=0\). Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) where \(a=-8\), \(b = 16\), \(c = 2\).

$$t=\frac{-16\pm\sqrt{16^{2}-4\times(-8)\times2}}{2\times(-8)}=\frac{-16\pm\sqrt{256 + 64}}{-16}=\frac{-16\pm\sqrt{320}}{-16}=\frac{-16\pm8\sqrt{5}}{-16}$$

$$t = 1\pm\frac{\sqrt{5}}{2}$$

We take the positive root \(t=1+\frac{\sqrt{5}}{2}\approx2.12\) (E).

Step2: Find the time to reach maximum height

For a quadratic function \(y = ax^{2}+bx + c\), the \(x\) - coordinate of the vertex (in this case \(t\)) is given by \(t=-\frac{b}{2a}\). Here \(a=-8\), \(b = 16\), so \(t=-\frac{16}{2\times(-8)} = 1\) (C).

Step3: Find the maximum height

Substitute \(t = 1\) into \(h(t)\): \(h(1)=-8\times1^{2}+16\times1+2=-8 + 16+2=10\) (A).

Step4: Find the height of the launch platform

Set \(t = 0\) in \(h(t)\), \(h(0)=-8\times0^{2}+16\times0+2=2\) (G).

Answer:

ECAG