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anonymous

  • 5 years ago

Algebra 2 Scott launches a model rocket from ground level. The rocket's height h in meters is given by the equation h = -4.9t sq + 56t, where t is the time in seconds after the launch. a) what is the maxium height the rocket will reach? b) how long after it is launched will the rocket reach its maxium height? c) Scott stands on a stool that is 3 meters off the ground to launch the rocket, what is the new height equation? What is the new max height?

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  1. anonymous
    • 5 years ago
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    hey is this a calc/derivative problem?

  2. anonymous
    • 5 years ago
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    or are you supposed to be solving it in a different way

  3. anonymous
    • 5 years ago
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    you can do this with the first derivative

  4. anonymous
    • 5 years ago
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    you want to know when the rate of change of the height is zero, so you take the first derivative

  5. anonymous
    • 5 years ago
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    1. h = -4.9t^2 + 56t 2. dh/dt = -4.9*2*t + 56

  6. anonymous
    • 5 years ago
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    3. 56 -9.8t

  7. anonymous
    • 5 years ago
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    there will be a maximum when that rate is zero

  8. anonymous
    • 5 years ago
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    so 56 = 9.8t

  9. anonymous
    • 5 years ago
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    t - 56/9.8

  10. anonymous
    • 5 years ago
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    errr t = 56/9.8

  11. anonymous
    • 5 years ago
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    = 560/98 = 280/49 = 40/7

  12. anonymous
    • 5 years ago
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    so at 40/7 seconds, it will reach its highest point

  13. anonymous
    • 5 years ago
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    so 40/7 seconds is the correct answer for (B). if you put that back into original height equation, you will have the answer to A.

  14. anonymous
    • 5 years ago
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    the answer to C is just 3 meters plus the answer to A

  15. anonymous
    • 5 years ago
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    since the rocket still peaks after the same amount of time - just three meters higher

  16. anonymous
    • 5 years ago
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    let me know if you were supposed to do this in a different way (eg. graphing)

  17. anonymous
    • 5 years ago
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    can you do it graphing? That would be a great help to have both methods.

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