Quadratic Equation:




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

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.


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
\(30\) 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
\(t=\frac{4}{5}+\frac{-\sqrt{30}}{10}\) \(t=\frac{4}{5}-\frac{-\sqrt{30}}{10}\)




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

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