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Homework Help: Difference quotient problem and average rate of change

  1. Jul 15, 2013 #1
    1. The problem statement, all variables and given/known data

    A soccer ball is kicked into the air from a platform. The height of the ball, in metres, t seconds after it is kicked is modelled by h(t) = -4.9t^2 + 13.5t + 1.2.

    I solved for the expression that represents the average rate of change of height over the interval 2<= t <= (2+h) and got which is correct (according to the answer key):

    [-4.9(2+h)^2 + 13.5(2+h) - 7.4] / h

    For what values of h is the expression above not valid?

    The answer in the back states h>0.84; at 2.84 s, h = 0.

    I understood how they got h>0.84 by finding the zeroes of the quadratic equation to get 2.84 and then subbing that into the inequality to get the change (h) to be 0.84. Therefore since t !> 2.84 h !> 0.84. But how do they get at 2.84 s, h != 0 as restrictions?

    How did they get 0.84 and 2.84 in the restrictions. This question is a bit confusing to me. Can anyone help me find the answer?
    Last edited: Jul 15, 2013
  2. jcsd
  3. Jul 15, 2013 #2


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    2017 Award

    Staff: Mentor

    It is confusing that h is both used as the height of the object and a time.

    h(t)=0 (the height) corresponds to the ball landing on the floor. This will certainly stop its free fall, and make the formula for h(t) invalid after that point.
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