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KJ4UTS

  • one year ago

Suppose a rock is thrown upwards from the top of a building 200 feet high with an initial velocity of 96 feet per second. The height (in feet) of the rock above the ground t seconds after being released is given by the formula h(t) = -16t^2 + 96t + 200 1.When will it hit the ground? 2.How high will it be after 2.1 seconds? 3.What is the maximum height?

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  1. anonymous
    • one year ago
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    1. It will hit the ground when h(t) = 0 so you need to solve \[-16t^2+96t+200=0\]

  2. anonymous
    • one year ago
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    try the quadratic formula

  3. KJ4UTS
    • one year ago
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    Would it be t=3-√ 43/2=-1.63 and 3+√ 43/2=7.63

  4. anonymous
    • one year ago
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    yes, and since time can't be negative, it hits the ground at 7.63 seconds.

  5. anonymous
    • one year ago
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    For 2, plug in 2.1 for t to find h(2.1).

  6. KJ4UTS
    • one year ago
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    331.04

  7. anonymous
    • one year ago
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    yes. and then for 3 there are two steps. Use \(t=-\frac{ b }{ 2a }\) to find the time of max height. Then plug that time into the h(t) equation

  8. KJ4UTS
    • one year ago
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    Im a little confused as to what I plug into b and a?

  9. anonymous
    • one year ago
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    the same values you used in the quadratic equation a = -16, b = 96, c = 200

  10. anonymous
    • one year ago
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    \[t=-\frac{ 96 }{ 2(-16) }\]

  11. KJ4UTS
    • one year ago
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    3

  12. anonymous
    • one year ago
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    now plug 3 in for t in the original equation

  13. KJ4UTS
    • one year ago
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    Ok I got 344 thank you for explaining this step by step :)

  14. anonymous
    • one year ago
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    yeah that's it. you're welcome. :)

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