- anonymous

s(t)=-16x^2+80t+100

- katieb

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- zepdrix

\[\large\rm s(t)=-16t^2+80t+100\]What's up Cheyenne? :)
What do you need to do with this function?

- anonymous

This is the problem :
1. A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. The path of the
ball is modeled by the function:
After how many seconds does the projectile reach its maximum height? What is this maximum height? After how
many seconds will the ball hit the ground?

- zepdrix

The shape of the path the ball follows is parabolic.
So the maximum height is going to be located at the `vertex` of the parabola.
Do you remember how to put a parabola into vertex form? :)

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## More answers

- anonymous

no I do not

- zepdrix

\[\large\rm -16t^2+80t\qquad\qquad +100\]Ignore the +100 for now.
We want to try and complete the square on the t's.
We'll start by factoring -16 out of each term to get the coefficient off of the squared term.\[\large\rm -16(\color{orangered}{t^2-5t})\qquad\qquad +100\]And now we want to try and complete the square on this orange part.

- zepdrix

In case there in any confusion on what I just did:
I divide -16 out of 16t^2, leaving t^2 in the first part of the bracket.
And I divided -16 out of 80 leaving a 5 on the t.
And the -16 is multiplying the outside of the brackets.

- zepdrix

So uhhh to complete the square :p let's see...

- anonymous

okay i get the dividing out the -16 part

- zepdrix

Let's split the t's into two equal groups, not the squares but the other ones.
\[\large\rm t^2-5t\quad=\quad t^2-2.5t-2.5t\]We can draw a square to show this relationship.
The side length will be t-2.5.|dw:1441252166003:dw|

- zepdrix

|dw:1441252302630:dw|If we fill in the area of the pieces, we can see that to complete the square we need the bottom right corner piece.
Which is -2.5 * -2.5 in area.

- anonymous

6.25 ?

- zepdrix

If the squares are confusing, we can just stick to the formulas :)
The formula tells you to take half of your b coefficient, and square it.
So our b value is -5,
half of that is -2.5, squaring gives us 6.25.
Good good good, that's the value that completes the square for us.

- zepdrix

\[\large\rm -16(\color{orangered}{t^2-5t+6.25})\qquad\qquad +100\]But we can't just add 6.25 willy nilly like this.
We'll keep the equation balanced by also subtracting 6.25.\[\large\rm -16(\color{orangered}{t^2-5t+6.25}-6.25)\qquad\qquad +100\]

- zepdrix

But the -6.25 isn't part of our perfect square, we want to take it outside of the brackets.
See how there is a -16 multiplying that entire bracket?
We have to multiply the value by -16 to take it outside.

- zepdrix

\[\large\rm =-16(\color{orangered}{t^2-5t+6.25})-16(-6.25)+100\]

- zepdrix

And our perfect square condenses down to: \(\large\rm \left(t+\frac{b}{2}\right)^2\)
So we end up with:\[\large\rm =-16(t+2.5)^2-16(-6.25)+100\]

- zepdrix

Then you need to combine the numbers on the outside to find the `y-coordinate of your vertex`.
\(\large\rm f(x)=(x-h)^2+k\)
Where `h` is the x-coordinate of your vertex
and `k` is the y-coordinate of your vertex.
The y-coordinate of your vertex corresponds to the height at that moment.

- zepdrix

What do you think Chey? :d too confusing?

- anonymous

Im kinda understanding it my teacher didnt really teach us the vertex from she just thought we knew it

- zepdrix

So at time t=2.5,
the ball will reach it's highest height of \(\large\rm -16(-6.25)+100\) feet.
So how high up is that? :)
Come on Chey! You can do the last calculation! :D

- anonymous

200 ft

- zepdrix

yay good job \c:/
That takes care of the first part.

- zepdrix

Oh actually that's the second question..
I kinda spilled the beans on the first part :P it's the x-coordinate of the vertex, 2.5 seconds.

- zepdrix

The third question asks,
After how many seconds will the ball hit the ground?
So when the ball is on the ground...
that corresponds to a height of zero... since it's in the ground.
And remember, the function, s(t), represents height.
\[\large\rm s(t)=-16t^2+80t+100\]So we let the height be zero.\[\large\rm 0=-16t^2+80t+100\]And we solve for t to figure out at what time this actually happens.

- zepdrix

Do you understand how to solve for t?
Try factoring, if that's too confusing,
you can always turn to your `quadratic function`.\[\large\rm t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

- zepdrix

quadratic formula*

- anonymous

so you would take out a -4 . so then it would be \[-4(4t^2-20t-25)\]

- zepdrix

Mmm ya you could do that :)\[\large\rm 0=-4(4t^2-20t-25)\]And then divide both sides by -4,\[\large\rm 0=4t^2-20t-25\]That certainly simplifies it a tad bit.

- anonymous

so then do i factor the \[0=(4t^2-20t-25)\]

- zepdrix

Oh it appears it won't factor :\
So let's save ourselves the trouble of even trying.
Just plug the numbers into your quadratic formula:\[\large\rm t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]and simplify! :)

- anonymous

so would the 20 be positive or negative for the -b

- zepdrix

\[\large\rm t=\frac{-(b)\pm\sqrt{(b)^2-4ac}}{2a}\]You're plugging a -20 into the b,\[\large\rm t=\frac{-(-20)\pm\sqrt{(-20)^2-4ac}}{2a}\]So I guess that first one is going to be positive, ya?

- anonymous

okay so then what ?

- zepdrix

Le sigh.. :)
Plug the other values in silly!\[\large\rm t=\frac{-(-20)\pm\sqrt{(-20)^2-4(4)(-25)}}{2(4)}\]And simplify! :D

- anonymous

I had already plugged all the values in :)

- anonymous

so the \[20^2 = 400 and the -4 *4*-25 equals 400\]

- anonymous

Do you mind helping with a different problem ?

- zepdrix

Ya so you get uhhhhh\[\large\rm t=\frac{-(-20)\pm\sqrt{400-(-100)}}{2(4)}\]\[\large\rm t=\frac{20\pm\sqrt{400+400}}{8}\]\[\large\rm t\approx\frac{20\pm28.28}{8}\]Where the positive root is the only one that gives us a positive t value.
\[\large\rm t\approx\frac{20+28.28}{8}\]Which gives you like 6ish.
something like this ya? :o

- zepdrix

A different problem? >.<
Grr I have so much homework to dooooo :O
Close down this thread, its getting too long and messy.
Open a new one with your new question.
You can type @zepdrix to page me, I'll come take a look if I have time :DDD

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