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h=-5t2+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
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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
 $$t=\frac{4}{5}+\frac{-\sqrt{30}}{10}$$ $$t=\frac{4}{5}-\frac{-\sqrt{30}}{10}$$

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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