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anonymous

  • one year ago

An object is launched upward at 45 ft/sec from a platform that is 40 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t2 + 45t + 40? 65 feet 72 feet 80 feet 93 feet

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  1. Mehek14
    • one year ago
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    what did you get?

  2. anonymous
    • one year ago
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    okay so, ik i can take a 4 out of all of them

  3. welshfella
    • one year ago
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    maximum height is when velocity = 0

  4. Mehek14
    • one year ago
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    use the same formula as last time

  5. anonymous
    • one year ago
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    okay -45/2(-16)

  6. Mehek14
    • one year ago
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    yes and then?

  7. anonymous
    • one year ago
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    -45/-32 1.4

  8. Mehek14
    • one year ago
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    it can actually be rounded to 1.41

  9. Mehek14
    • one year ago
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    now plug 1.41 into the equation

  10. anonymous
    • one year ago
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    okay then plug it in

  11. Mehek14
    • one year ago
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    yes

  12. anonymous
    • one year ago
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    i got 98

  13. Mehek14
    • one year ago
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    \(h(t) = -16*1.41^2 + 45*1.41 + 40\\ 1.41^2=1.9881*-16=-31.81\\45*1.41=63.45\\63.45-31.81=31.64\\31.64+40=71.64\) 71.64 can be rounded to 72

  14. anonymous
    • one year ago
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    ohh okay.

  15. anonymous
    • one year ago
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    thank you!

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