Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Graphing Problem?

Sarah is a world-class diver in the women’s 3-m springboard competition. Her height, h (in metres), above the water t seconds after she leaves the board is given h=4.9t^2+8.8t+3 by . How long is Sarah in the air before she reaches the water? Make sure to round your answer to the nearest tenth. Hint: Try graphing the problem on your calculator.

3 Answers

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  • 1 month ago

    In order to simulate a real jump, the function should be:

    h(t) = -4.9t² + 8.8t + 3

    She reaches the water when h(t) = 0

    i.e. when -4.9t² + 8.8t + 3 = 0

    Using the quadratic formula we have:

    t = [-8.8 ± √(8.8² - 4(-4.9)(3)]/-9.8

    so, t = (-8.8 ± √136.24)/-9.8

    Hence, t = 2.1 seconds

    :)>

  • Ash
    Lv 7
    1 month ago

    Pope is correct about the error. You missed the negative sign for first term 

    h = -4.9t²+8.8t+3

    When Sarah reaches the water h = 0

    0 = -4.9t²+8.8t+3

    Using quadratic solution

    t = {-8.8±√[8.8² - 4(-4.9)(3)]}/(2*-4.9)

    t = 2.1 s

    Attachment image
  • Pope
    Lv 7
    1 month ago

    Yes, do try graphing it on your calculator. Are you quite certain of the formula you gave for height? Once she leaves the board, she will continue to rise, never to return to Earth.

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