1. Re: Quadratic Formula Question #6

You shouldn't be guessing anything, but yes, h = 2 when t = 0.

As for c), I would think about the symmetry of the quadratic, which means that everywhere except the turning point, there are always two t-values for every h value, and due to the symmetry, the turning point has to be located exactly halfway between them.

You know that when t = 0, h = 2. Can you find another t-value for when h = 2? When you have that, average them and that gives you the t-value of the turning point. When you have the t-value of the turning point, substitute it into the equation to find the h-value of the turning point.

2. Re: Quadratic Formula Question #6

i also tried doing f again...

f) (20)t = -4.9t^2 + 32t + 2

t = -.245 +1.6t + .1

3. Re: Quadratic Formula Question #6

It is so hard to help you when you jump from one question to another. Please stick with one question at a time! We have established a) and b) are correct now, and I've given you some help with part c). Please attempt it and let's get that right before moving on!

4. Re: Quadratic Formula Question #6

b) t = 6.59 s this is correct? what about d)

i did not understand what you said about c) if you are saying there has to be 2 answers for 1 h, then are you just saying i need to use the quadratic formula? but you are talking about symettry, so do i just find graph the equation and find the maximum?

5. Re: Quadratic Formula Question #6

i'm sorry, let's stick with c) for now, haha

6. Re: Quadratic Formula Question #6

Do you understand what a parabola (quadratic graph) looks like? Can you see that if you drew a horizontal line through it, you will have TWO intersections? That means that there will be two values of t for each h (except at the turning point where there is only one)?

Can you see how the graph is symmetrical? Do you understand then that if you have two values of t with the same h, because the graph is symmetrical, that the turning point's t-value will be right in the middle of them?

7. Re: Quadratic Formula Question #6

yes, i understand that now, so i got the maximum from the graph and it is where the axis of symmetry is. (3.27, 54.2)

so is it, 54.2 m high in 3.27 s?

8. Re: Quadratic Formula Question #6

I think I have actually figured out how to do all the questions except e)

All I have done is converting 60 km/h to m/s

= 16.66 m/s

Am I suppose to make up an equation myself?

9. Re: Quadratic Formula Question #6

For part e), you want to equate $\displaystyle h(t)=0$ and solve for $\displaystyle t$, discarding the negative root. This is what you did in your original post, however, your value for $\displaystyle t$ is incorrect. Once you correctly determine the length of time the disk is in the air, then, ignoring drag, you assume the horizontal velocity is constant, and use the relation between distance speed and time:

$\displaystyle d=vt$

to find the range. I have interpreted the question to mean distance traveled is the range, rather than the parabolic arc length for obvious reasons.

10. Re: Quadratic Formula Question #6

0 = -4.9t2 + 32t + 2

-4.9t + 32 + 2

-4.9t + 34

-35/-4.9 = t

t = 7.142857143 s

d = v / t

d = 16.66 / 7.14...

d = 2.33...

11. Re: Quadratic Formula Question #6

No, you need to use the quadratic formula (or complete the square). You cannot just reduce the degree of the variable like that.

We have:

$\displaystyle -4.9t^2 + 32t + 2=0$

Now, in the quadratic formula, let $\displaystyle a=-4.9,\,b=32,\,c=2$ and discard the negative root. What do you find?

12. Re: Quadratic Formula Question #6

I got x = -0.06 and x = 6.59,

is 6.59 the t?

13. Re: Quadratic Formula Question #6

Yes, to two decimal places, that is the positive root you are seeking.

14. Re: Quadratic Formula Question #6

If you plug in t = 4 then you get using R:

> -4.9*4^2 + 32*4 + 2
[1] 51.6

or 51.6 metres

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