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anonymous

  • one year ago

im new hi (moderators delete this please :) and my other question thanks)

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  1. zzr0ck3r
    • one year ago
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    well they tell you that v = 50 So what does the equation look like now?

  2. zzr0ck3r
    • one year ago
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    ok then they tell us that it starts off at 90 What do you get at the start of time? I.e. what do you get when \(t=0\)?

  3. zzr0ck3r
    • one year ago
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    put in \(t=0\)

  4. zzr0ck3r
    • one year ago
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    nono, so if we let \(t=0\) into the function we get \(s\) correct?

  5. zzr0ck3r
    • one year ago
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    right, and they told us that the intial height is 90, so s=90. make sense?

  6. zzr0ck3r
    • one year ago
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    so what is the function?

  7. zzr0ck3r
    • one year ago
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    nono, ok before we can start to answer the question we had to figure out what \(v\) and \(s\) are. Now we have done that and we have our function is \[H(t) -16t^2+50t+90\] Are you with me?

  8. zzr0ck3r
    • one year ago
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    Now we can start to actually answer the question.

  9. zzr0ck3r
    • one year ago
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    Ok part A wants to know \(t\) when the thing hits the ground. What is the height when something hits the ground?

  10. zzr0ck3r
    • one year ago
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    ok so we need to solve \[-16t^2+50t+90=0\] Can you do that?

  11. zzr0ck3r
    • one year ago
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    I am not sure if that is correct, but you can check that. Ok notice that since we are dealing with time, that we would only take the positive answer. That is part A. For part B, do you know how to find the vertex?

  12. zzr0ck3r
    • one year ago
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    the \(x\) coordinate of the vertex of the function \(ax^2+bx+c\) is \(x=\dfrac{-b}{2a}\).

  13. zzr0ck3r
    • one year ago
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    what do you get?

  14. zzr0ck3r
    • one year ago
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    what is \(a\) and \(b\) in your equation? \(ax^2+bx+c\)?

  15. zzr0ck3r
    • one year ago
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    nono \(-16x^2+50x+90\) \( \ \ \ ax^2+bx+c\) What is \(a\)?

  16. zzr0ck3r
    • one year ago
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    correct, ok so what is \(\dfrac{-b}{2a}\)?

  17. zzr0ck3r
    • one year ago
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    2*-16 = 100?

  18. zzr0ck3r
    • one year ago
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    correct, now plug that into the equation to get the height.

  19. zzr0ck3r
    • one year ago
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    I am not responsible for you failing a class. I also highly doubt that getting help on one question will change you from a failing grade to passing.

  20. zzr0ck3r
    • one year ago
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    so plug in -25/16 into the function and that will give you the max height.

  21. zzr0ck3r
    • one year ago
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    err 25/16

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