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Derivative question (Help Please)

  1. May 25, 2005 #1
    A dart is shot straight up from 1.5m above ground level. The distance d,
    in meters, the dart is above ground at time t, in seconds, is :

    d(t) = -4.9t^2 + 20t + 1.5

    For how many seconds is the dart in the air?
     
  2. jcsd
  3. May 25, 2005 #2

    arildno

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    What is the height value corresponding to ground level?
     
  4. May 25, 2005 #3
    There was a previous part to the question which I have the answer for :

    What is the velocity of the dart one second after it is shot upward?

    v'(t) = -9.8t + 20 = -9.8(1) + 20 = 10.2 m/s

    What is the Maximum height that the dart reaches ?

    velocity will be zero at 2.0408 seconds [0=(-9.8)t + 20]

    d(2.0408) = -4.9(2.0408)^2 + 20(2.0408) + 1.5 = 21.9 m
     
  5. May 25, 2005 #4

    arildno

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    Both of these earlier answers are correct.
     
  6. May 25, 2005 #5
    I am stuck on the final part of the question, I know how to solve it using a physics formula . . . but not with the derivative.
     
  7. May 25, 2005 #6

    arildno

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    Did you read post 2?
     
  8. May 25, 2005 #7
    It would be zero.
     
  9. May 25, 2005 #8

    arildno

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    Correct.
    So what equation must the instant "t" fulfill which coresponds to the dart hitting the ground?
     
  10. May 25, 2005 #9
    I'm guessing that something needs to be set to equal zero . . . but forgive me, I'm still a little lost.
     
  11. May 25, 2005 #10

    arildno

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    You know that height as a function of time is given by the equation:
    [tex]d(t)=-4.9t^{2}+20t+1.5 (1)[/tex]
    Now, that d=0 (that is, we have reached groundlevel), means that the instant "t" corresponding to that must fulfill:
    [tex]0=-4.9t^{2}+20t+1.5 (2)[/tex]

    Note that we have been sloppy with our notation here:
    In (1), "t" is used as a VARIABLE, whereas in (2), "t" is used as a fixed VALUE we're supposed to find.
    If you want to be careful in your notation, proceed as follows:
    a) Let T be the instant when the dart hits the ground. At that instant, the height value "d" is 0.
    b) The value of the height at time T is given by evaluating our height formula at T (T is therefore an element in the interval over which the variable "t" ranges):
    [tex]d(T)=-4.9T^{2}+20T+1.5 (3)[/tex]
    c) a) says now that d(T)=0, and inserting this insight into (3) yields:
    [tex]0=d(T)=-4.9T^{2}+20T+1.5[/tex]
    That is:
    [tex]0=-4.9T^{2}+20T+1.5(4)[/tex]

    (4) can now be solved for T, remembering that T must be greater than zero.
     
    Last edited: May 25, 2005
  12. May 25, 2005 #11
    Thanks for your help today. :smile:
     
  13. May 25, 2005 #12

    arildno

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    Welcome to PF!
     
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