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anonymous
 one year ago
s(t)=16x^2+80t+100
anonymous
 one year ago
s(t)=16x^2+80t+100

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4\[\large\rm s(t)=16t^2+80t+100\]What's up Cheyenne? :) What do you need to do with this function?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This 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
 one year ago
Best ResponseYou've already chosen the best response.4The 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? :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4\[\large\rm 16t^2+80t\qquad\qquad +100\]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.\[\large\rm 16(\color{orangered}{t^25t})\qquad\qquad +100\]And now we want to try and complete the square on this orange part.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4In 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
 one year ago
Best ResponseYou've already chosen the best response.4So uhhh to complete the square :p let's see...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay i get the dividing out the 16 part

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Let's split the t's into two equal groups, not the squares but the other ones. \[\large\rm t^25t\quad=\quad t^22.5t2.5t\]We can draw a square to show this relationship. The side length will be t2.5.dw:1441252166003:dw

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4dw:1441252302630:dwIf 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4If 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
 one year ago
Best ResponseYou've already chosen the best response.4\[\large\rm 16(\color{orangered}{t^25t+6.25})\qquad\qquad +100\]But we can't just add 6.25 willy nilly like this. We'll keep the equation balanced by also subtracting 6.25.\[\large\rm 16(\color{orangered}{t^25t+6.25}6.25)\qquad\qquad +100\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4But 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
 one year ago
Best ResponseYou've already chosen the best response.4\[\large\rm =16(\color{orangered}{t^25t+6.25})16(6.25)+100\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4And our perfect square condenses down to: \(\large\rm \left(t+\frac{b}{2}\right)^2\) So we end up with:\[\large\rm =16(t+2.5)^216(6.25)+100\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Then you need to combine the numbers on the outside to find the `ycoordinate of your vertex`. \(\large\rm f(x)=(xh)^2+k\) Where `h` is the xcoordinate of your vertex and `k` is the ycoordinate of your vertex. The ycoordinate of your vertex corresponds to the height at that moment.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4What do you think Chey? :d too confusing?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im kinda understanding it my teacher didnt really teach us the vertex from she just thought we knew it

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4So 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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4yay good job \c:/ That takes care of the first part.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Oh actually that's the second question.. I kinda spilled the beans on the first part :P it's the xcoordinate of the vertex, 2.5 seconds.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4The 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
 one year ago
Best ResponseYou've already chosen the best response.4Do you understand how to solve for t? Try factoring, if that's too confusing, you can always turn to your `quadratic function`.\[\large\rm t=\frac{b\pm\sqrt{b^24ac}}{2a}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you would take out a 4 . so then it would be \[4(4t^220t25)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Mmm ya you could do that :)\[\large\rm 0=4(4t^220t25)\]And then divide both sides by 4,\[\large\rm 0=4t^220t25\]That certainly simplifies it a tad bit.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so then do i factor the \[0=(4t^220t25)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Oh it appears it won't factor :\ So let's save ourselves the trouble of even trying. Just plug the numbers into your quadratic formula:\[\large\rm t=\frac{b\pm\sqrt{b^24ac}}{2a}\]and simplify! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so would the 20 be positive or negative for the b

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4\[\large\rm t=\frac{(b)\pm\sqrt{(b)^24ac}}{2a}\]You're plugging a 20 into the b,\[\large\rm t=\frac{(20)\pm\sqrt{(20)^24ac}}{2a}\]So I guess that first one is going to be positive, ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Le sigh.. :) Plug the other values in silly!\[\large\rm t=\frac{(20)\pm\sqrt{(20)^24(4)(25)}}{2(4)}\]And simplify! :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I had already plugged all the values in :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the \[20^2 = 400 and the 4 *4*25 equals 400\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you mind helping with a different problem ?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Ya 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
 one year ago
Best ResponseYou've already chosen the best response.4A 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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