4. At the end of the school year, Rachel and Amber threw their Algebra terbooks off the top of a 12-story building. The equation of the

Question

4. At the end of the school year, Rachel and Amber threw their Algebra terbooks off the top of a 12-story
building. The equation of the pathway of each book is given below. By how many seconds does Rachel’s
textbook beat Amber’s textbook to the ground?
Rachel: h= -16° + 367 + 160
Ambersh=-164 +501 + 150​

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Evelyn 2 weeks 2021-11-25T08:16:03+00:00 1 Answer 0

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    2021-11-25T08:17:26+00:00

    I’m going to decode the gibberish in the question and guess the equations are

    Rachel:

    h=-16t^2+36t+160

    Amber:

    h=-16t^2+50t+160

    These equations tell us a few things not explicitly mentioned in the problem.  Namely h is in feet, t is in seconds (because of the 16) and the building is 160 feet tall.  h=0 corresponds to ground level.  I made the constants of the equations the same because they’re both on the twelfth floor.  The difference between 36t and 50t reflect different vertical components of the velocity of the throw. The positive signs mean they each threw their book upwards.  Air resistance is ignored.

    So we need to solve two quadratic equations and compare the t values.

    0=-16t^2+36t+160

    t = \dfrac{-36 \pm \sqrt{36^2 - 4(-16)(160)}}{2(-16)}

    t = \dfrac{-36 \pm \sqrt{16(721)}}{-32}

    t = \dfrac{9 \pm\sqrt{721}}{8}

    Only the plus sign gives a positive value of t, so that’s our answer for Rachel:

    t_R = \dfrac{9+ \sqrt{721}}{8} \approx 4.48

    For Amber we get

    0=-16t^2+50t+160

    t = \dfrac{-50 \pm \sqrt{50^2 - 4(-16)(160)}}{2(-16)}

    t = \dfrac{-50 \pm \sqrt{4(49)(65)}}{-32}

    t = \dfrac{25 \pm 7\sqrt{65}}{16}

    Again we choose the plus sign,

    t_A = \dfrac{25+7\sqrt{65}}{16} \approx 5.09

    Rachel beat Amber by

    t = t_A - t_R = 5.09-4.58 = 0.51

    Answer: 0.51 seconds

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