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anonymous

  • one year ago

will fan & medal!!!!!!!!!!!!!! 2. A model rocket is launched from the ground with an initial velocity of 288 ft/sec. c. How long will it take the rocket to reach its maximum height? Show all work in the space provided.

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  1. misty1212
    • one year ago
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    HI!!

  2. misty1212
    • one year ago
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    i am guessing you are supposed to use \[h(t)=-16t^2+228t\]

  3. misty1212
    • one year ago
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    max is at the vertex first coordinate of the vertex is always \[-\frac{b}{2a}\] which in your case is \[-\frac{288}{2\times (-16)}\]

  4. anonymous
    • one year ago
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    d. Assume the model rocket’s parachute failed to deploy and the rocket fell back to the ground. How long would it take the rocket to return to Earth from the time it was launched? Show all work in the space provided.

  5. misty1212
    • one year ago
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    set \[-16t^2+288t=0\] and solve for \(t\) takes 3 steps only

  6. misty1212
    • one year ago
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    or maybe two

  7. misty1212
    • one year ago
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    factors nicely as \[ -16t (t-18) = 0\]

  8. anonymous
    • one year ago
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    another way to think about it is to simply take the derivative of the equation 'misty1212' posted which is -32t=-288 and t=288/32. to prove this, the second derivative is -32 which is < 0 therefore it is a maximum.

  9. anonymous
    • one year ago
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    thnx a lot guys really appreciate the help

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