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anonymous

  • 4 years ago

why is this question so difficult? The height of a rocket launched from the ground is given by the function s(t) = -16 t^2 + 80 t, where s is in feet above ground and t is in seconds. After 3 seconds, the rocket is ________ feet above ground.

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  1. anonymous
    • 4 years ago
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    then After 3 seconds, the rocket is traveling at a velocity of _________ feet per second.

  2. lgbasallote
    • 4 years ago
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    Can I just answer your first question? Because it is..lol xD

  3. anonymous
    • 4 years ago
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    lol sure!!

  4. anonymous
    • 4 years ago
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    s(3)=-16(3)^2+80(3)=96 ft

  5. anonymous
    • 4 years ago
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    v(t)=s'(t)=-32t+80 v(3)=-32(3)+80=-16ft/s

  6. anonymous
    • 4 years ago
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    thanks

  7. anonymous
    • 4 years ago
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    wait how do I find if The rocket is 60 feet above ground at what times?

  8. anonymous
    • 4 years ago
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    set s(t)=60,solve for t

  9. anonymous
    • 4 years ago
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    you'll have a quadratic equation, it's a parabola so there should be 2 answers

  10. anonymous
    • 4 years ago
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    ok i got -38,200

  11. anonymous
    • 4 years ago
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    and the the other answer is pos. 38,200

  12. anonymous
    • 4 years ago
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    am I right?

  13. anonymous
    • 4 years ago
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    am I right?

  14. anonymous
    • 4 years ago
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    let me check

  15. anonymous
    • 4 years ago
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    i highly doubt those are the right answers, the rocket will have hit the ground long before those times

  16. anonymous
    • 4 years ago
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    \[-16t^2+80x-60=0\] Solving for t, \[t= \frac{(5 \pm \sqrt{10})}{2}\]

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