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anonymous

  • one year ago

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

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

  2. anonymous
    • one year ago
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    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?

  3. zepdrix
    • one year ago
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    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? :)

  4. anonymous
    • one year ago
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    no I do not

  5. zepdrix
    • one year ago
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    \[\large\rm -16t^2+80t\qquad\qquad +100\]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.

  6. zepdrix
    • one year ago
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    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.

  7. zepdrix
    • one year ago
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    So uhhh to complete the square :p let's see...

  8. anonymous
    • one year ago
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    okay i get the dividing out the -16 part

  9. zepdrix
    • one year ago
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    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|

  10. zepdrix
    • one year ago
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    |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.

  11. anonymous
    • one year ago
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    6.25 ?

  12. zepdrix
    • one year ago
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    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.

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

  14. zepdrix
    • one year ago
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    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.

  15. zepdrix
    • one year ago
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    \[\large\rm =-16(\color{orangered}{t^2-5t+6.25})-16(-6.25)+100\]

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

  17. zepdrix
    • one year ago
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    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.

  18. zepdrix
    • one year ago
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    What do you think Chey? :d too confusing?

  19. anonymous
    • one year ago
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    Im kinda understanding it my teacher didnt really teach us the vertex from she just thought we knew it

  20. zepdrix
    • one year ago
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    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

  21. anonymous
    • one year ago
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    200 ft

  22. zepdrix
    • one year ago
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    yay good job \c:/ That takes care of the first part.

  23. zepdrix
    • one year ago
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    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.

  24. zepdrix
    • one year ago
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    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.

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

  26. zepdrix
    • one year ago
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    quadratic formula*

  27. anonymous
    • one year ago
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    so you would take out a -4 . so then it would be \[-4(4t^2-20t-25)\]

  28. zepdrix
    • one year ago
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    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.

  29. anonymous
    • one year ago
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    so then do i factor the \[0=(4t^2-20t-25)\]

  30. zepdrix
    • one year ago
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    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! :)

  31. anonymous
    • one year ago
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    so would the 20 be positive or negative for the -b

  32. zepdrix
    • one year ago
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    \[\large\rm t=\frac{-(b)\pm\sqrt{(b)^2-4ac}}{2a}\]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?

  33. anonymous
    • one year ago
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    okay so then what ?

  34. zepdrix
    • one year ago
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    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

  35. anonymous
    • one year ago
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    I had already plugged all the values in :)

  36. anonymous
    • one year ago
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    so the \[20^2 = 400 and the -4 *4*-25 equals 400\]

  37. anonymous
    • one year ago
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    Do you mind helping with a different problem ?

  38. zepdrix
    • one year ago
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    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

  39. zepdrix
    • one year ago
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    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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