Pablo and max's follows a flight pattern approximatley modled by the equation h(t)=-8t^2+16t+2 where h(t)is their rocket's height in meters t seconds after launch. How many seconds was their rocket in the air

Answer: 2.11 s

Step-by-step explanation:

Given

Equation of rockets height is given by

[tex]\Rightarrow h(t)=-8t^2+16+2[/tex]

Height will be zero once the rocket lands on the ground

So, we can write

[tex]\Rightarrow h(t)=0\ \text{or}\ -8t^2+16t+2=0\\\Rightarrow 8t^2-16t-2=0\\\\\Rightarrow t=\dfrac{16\pm \sqrt{(-16)^2-4(8)(-2)}}{2\times 8}\\\\\Rightarrow t=\dfrac{16\pm \sqrt{320}}{16}\\\\\text{Neglecting negative value of t}\\\\\Rightarrow t=\dfrac{16+17.888}{16}=2.11\ s[/tex]

The rocket of height h(t) and modeled equation [tex]\rm h(t)=-8t^2+16t+2\\[/tex]

was in the air for 2.1 seconds.

The given equation is  [tex]\rm h(t)=-8t^2+16t+2\\[/tex], where h(t) is the rocket's height in meters t seconds after launch.

In algebra, a quadratic equation is an equation that can be written in standard form as

[tex]\rm ax^{2}+bx+c=0}[/tex]

where x represents variable, and a, b, and c represent coefficient.

The given equation is

h(t) = 0

[tex]\rm h(t)=-8t^2+16t+2\\h(t)=-8t^2+16t+2 = 0\\[/tex][tex]\rm h(t)=-8t^2+16t+2\\[/tex]

The roots will be

[tex]\rm {\displaystyle t={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}[/tex]

[tex]\rm {\displaystyle t={\frac {16\pm {\sqrt {(-16)^{2}-4\times 8\times 2}}}{16}}.}\\\rm {\displaystyle t={\frac {16\pm {\sqrt{320} }}{16}}}\\\rm {\displaystyle t={\frac{16+17.88}{16}\\[/tex]

Here, we have neglected the negative value of t

t = 2.1 seconds

So, the rocket of height h(t) and modeled equation [tex]\rm h(t)=-8t^2+16t+2\\[/tex]

was in the air for 2.1 seconds.

Here learn more about quadratic equations here:

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