We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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