Quadratic Equation:

h=-5t^{2}+8t+1.3

Where h=height and t=time

When does it first reach a high of 3m?

Guest Aug 6, 2017

edited by
Guest
Aug 6, 2017

#1**0 **

I am assuming that you want this equation solved using the quadratic formula. I would use that method, too. Of course, let's remind ourselves of the quadratic formula:

In a quadratic function in the form \(ax^2+bx+c=0\).

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

The question, however, asks for how much time elapsed when it (I assume a thrown ball) reached a height of 3 meters

\(3=-5t^2+8t+1.3\) | Subtract 3 on both sides. |

\(0=-5t^2+8t-1.7\) | |

Now that the equation is in the form of \(ax^2+bx+c=0\), let's solve for how much time, *t*, it took for it to reach 3 meters:

\(0=-5t^2+8t-1.7\) | Use the quadratic formula to solve for t. | ||

\(t = {-8 \pm \sqrt{8^2-4(-5)(-1.7)} \over 2(-5)}\) | We should first calculate the discriminant \(b^2-4ac\) to see if there are indeed solutions for t. Let's just focus on that part. | ||

\(8^2-4(-5)(-1.7)\) | Do 8^2 first. | ||

\(64-4(-5)(-1.7)\) | Do 4*-5 next. | ||

\(64-(-20)(-1.7)\) | Do -20*-1.7 | ||

\(64-34\) | |||

\(30\) | Because \(b^2-4ac>0\), this means that there are 2 solutions. Replace \(8^2-4(-5)(-1.7)\) with its calculated value, 30. | ||

\(t=\frac{-8\pm\sqrt{30}}{2(-5)}\) | SImplify 2*-5 in the denominator | ||

\(t=\frac{-8\pm\sqrt{30}}{-10}\) | Break up the fraction by doing \(\frac{a\pm b}{c}=\frac{a}{c}\pm\frac{b}{c}\) | ||

\(t=\frac{-8}{-10}\pm\frac{\sqrt{30}}{-10}\) | Simplify -8/-10 by realizing its GCF is -2. | ||

\(t=\frac{4}{5}\pm\frac{-\sqrt{30}}{10}\) | Split your answer into two answers | ||

| |||

The question asks for the first time that the ball hits 3 meters. Of course, we do not know the exact value of either of our solutions currently. We can utilize logical thinking to figure it out, however.

If we pretend as if \(\sqrt{30}=5\), as that is an OK approximation, we can simplify both solutions into

\(t=\frac{4}{5}-\frac{1}{2}\) and \(t=\frac{4}{5}+\frac{1}{2}\)

We can infer that \(\frac{4}{5}-\frac{1}{2}\) is still positive but closer to 0 because 4/5 is greater than 1/2. Subtracting the two would make it closer to 0. Adding both would make the numbers further from 0.

Therefore, the first time it goes to the height of 3 meters is after \(\frac{4}{5}-\frac{\sqrt{30}}{10}\) that much time elapsed. \(\frac{4}{5}-\frac{\sqrt{30}}{10}\approx0.252277\).

I have supllied a link to a graph if ou would like to check it out: https://www.desmos.com/calculator/ivnsia7gdd

TheXSquaredFactor
Aug 6, 2017