Question

1. the formula for the height of a bottle rocket launched up into the air from a height of 2 ft with a speed of 120 ft/sec is $h(t) = -16t^2 + 120t + 2$. how many seconds after it is launched will it take for the rocket to be at a height of 200 feet above the ground?

Explanation:

Step1: Set up the equation

We know the height function \( h(t)= - 16t^{2}+120t + 2\), and we want to find \( t\) when \( h(t) = 200\). So we set up the equation:
\(-16t^{2}+120t + 2=200\)

Step2: Simplify the equation

Subtract 200 from both sides of the equation:
\(-16t^{2}+120t+2 - 200=0\)
\(-16t^{2}+120t - 198 = 0\)
Multiply both sides by - 1 to make the coefficient of \(t^{2}\) positive:
\(16t^{2}-120t + 198=0\)
We can simplify this equation by dividing all terms by 2:
\(8t^{2}-60t + 99 = 0\)

Step3: Use the quadratic formula

For a quadratic equation \(ax^{2}+bx + c = 0\), the quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 8\), \(b=-60\), and \(c = 99\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-60)^{2}-4\times8\times99=3600 - 3168 = 432\)
Then, find the square root of the discriminant: \(\sqrt{432}=\sqrt{144\times3}=12\sqrt{3}\approx12\times1.732 = 20.784\)
Now, substitute into the quadratic formula:
\(t=\frac{60\pm20.784}{2\times8}=\frac{60\pm20.784}{16}\)
We have two solutions:
\(t_{1}=\frac{60 + 20.784}{16}=\frac{80.784}{16}\approx5.05\)
\(t_{2}=\frac{60-20.784}{16}=\frac{39.216}{16}\approx2.45\)
Since the rocket is going up and then coming down, both times are valid, but we can check which one makes sense in the context. When the rocket is launched, it goes up, reaches a maximum height, and then comes down. So there are two times when it is at 200 feet, once on the way up and once on the way down.

Answer:

The rocket will be at 200 feet at approximately \(t\approx2.45\) seconds (on the way up) and \(t\approx5.05\) seconds (on the way down). If we consider the first time it reaches 200 feet (on the ascent), the answer is approximately \(2.45\) seconds, and if we consider both, the times are approximately \(2.45\) seconds and \(5.05\) seconds. (If we solve the quadratic equation \(8t^{2}-60t + 99 = 0\) using the quadratic formula more precisely:

\(t=\frac{60\pm\sqrt{(-60)^{2}-4\times8\times99}}{2\times8}=\frac{60\pm\sqrt{3600 - 3168}}{16}=\frac{60\pm\sqrt{432}}{16}=\frac{60\pm12\sqrt{3}}{16}=\frac{15\pm3\sqrt{3}}{4}\)

\(\frac{15 + 3\sqrt{3}}{4}\approx\frac{15+5.196}{4}=\frac{20.196}{4}\approx5.05\)

\(\frac{15 - 3\sqrt{3}}{4}\approx\frac{15 - 5.196}{4}=\frac{9.804}{4}\approx2.45\))