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Accelaration and Free-fall problems

  1. Sep 17, 2009 #1
    Question A:

    This one is easy but my answer is coming out wrong, maybe my signs are wrong. The question is as follows:

    A ball is launched directly upward from ground level with an initial speed of 19 m/s.

    How many seconds after launch is the ball 8 m above the release point?

    I set it up like this but it keeps coming wrong:

    -8 m = 19t - 1/2 * 9.8 * t2(squared)

    Question B:

    This one is driving me nuts, I swear I have it but I dunno.

    A typical automobile under hard braking loses speed at a rate of about 6.7 m/s2; the typical reaction time to engage the brakes is 0.45 s. A local school board sets the speed limit in a school zone such that all cars should be able to stop in 3.8 m.

    (a) What maximum speed does this imply for an automobile in this zone?

    I used V2 = V02 + 2ad and it looks like:

    0 = v02 + 2 * -6.7 * 3.8

    The answer is in m/s, convert that to mi/h to get 15.98 mi/h. The other part to the question is:

    (b) What fraction of the 3.8 m is due to the reaction time?

    Which I believe I need part 1 for.

    Question C:

    This question I have no clue where to start.

    At t = 0, a stone is dropped from a cliff above a lake; 2.2 seconds later another stone is thrown downward from the same point with an initial speed of 49 m/s. Both stones hit the water at the same instant. Find the height of the cliff.

    Any help at all will be appreciated greatly...
  2. jcsd
  3. Sep 17, 2009 #2


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    Welcome to PF!

    Hi Raheelp! Welcome to PF! :smile:
    If up is positive, then it should be +8, not -8 :wink:

    You need two equations, one for the reaction time (zero acceleration), and one for the braking. :wink:
    Show us what equations you have got, and then we'll know how to help! :smile:
  4. Sep 18, 2009 #3
    Re: Welcome to PF!

    OK ty on question A it was obvious..

    Question B im trying right now.

    The equations we use are the usual three of motion and utilize average speed, average velocity.
  5. Sep 18, 2009 #4
    So for B:

    3.8 m = segment 1 d + segment 2 d ?
  6. Sep 18, 2009 #5


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    Yup! :biggrin:

    Show us what you get. :smile:
  7. Sep 18, 2009 #6
    Yeah I can't do B, I fail at this.

    So far I have this:

    For segment 1 I have

    a = 0
    t = .45
    d = x
    v0 = ?
    v = ?

    Segment 2

    a = -6.7
    t = ?
    d = 3.8 - x
    v0 = ?
    v = 0

    Do I solve for v0 in segment two, and put that in segment 1 to find d... ?
    Last edited: Sep 18, 2009
  8. Sep 18, 2009 #7


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    Yes. :smile:

    (but you can't call both of the distances 3.8 - x, can you? make the first distance x :wink:)

    (oh, and write vi if you're writing vf … write v0 only if you're writing v1 :smile:)
  9. Sep 18, 2009 #8
    Now I'm confused with what formula I can use.

    I think I'll have to get walked through the problem at the workshops on Monday.

    Thanks guys... Sorry I couldnt do anything with your expertise.
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