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Rock is being thrown horizontally off a cliff (2d motion)

  1. Feb 21, 2017 #1
    1. The problem statement, all variables and given/known data
    A rock is thrown horizontally off a 56 m high cliff overlooking the ocean. And the sound of the splash is heard 3.60 seconds later. Speed of sound in air is 343 m/s.

    What was the initial velocity of the rock?

    2. Relevant equations



    3. The attempt at a solution
    Okay so I know

    height of cliff is [itex]56m [/itex]

    [itex] V_0y = 0, V_x = V_{0x} [/itex]

    ##t = t_1 + t_2=3.60 s##

    Time it took for the rock to hit the ground I use this eq because ##V_oy = 0##

    ##\Delta y = V_{0y} t + \frac 12 a_x t^2##

    ## \sqrt {{2\Delta y }\over{a_x}} = t_1##

    plugging my values in I got

    ##t_1 = 3.38## whichc gives ##t_2 = 0.22 ##

    Since the cliff is 56 m high, I don't know the horizontal distance, but I do know that the speed of sound is traveling back in a straight line at ##343 m/s##, and with this, I multiply ##343 m/s## by ##0.22 s##, and get ##75 m##. This helps me get horizontal distance of ##50 m##. Does everyone agree with my logic here?

    So now I use formula
    since ##V_{0x} = V_x##

    ##\Delta x = {(V_{0x} + V_x)t\over 2}##

    ##{{\Delta x}\over t} = V_x##

    plugging in I get ##14.8 m/s ##

    But my book gives me answer ##15.1 m/s##

    Now I believe this to be an error that has to do with rounding or significant figures, but I'm not sure, and I don't see my professor until Saturday so I can't ask him if my method is correct. Can anyone please help me here?
     
    Last edited: Feb 21, 2017
  2. jcsd
  3. Feb 21, 2017 #2

    kuruman

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    Your method looks correct. It's probably round offs. Do it symbolically and plug in numbers at the very end to confirm this.
     
  4. Feb 21, 2017 #3

    gneill

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    Logic's okay, but you need to keep more digits in your intermediate values to avoid accumulated roundoff/truncation errors from creeping into your significant figures. Also, you haven't mentioned anywhere what value you are using for g.
    Not sure why you went to the trouble of using the formula you did. Since the x-velocity is constant you could simply write ##Δx = v_x t##. You should make clear which value of t you used, since there are three values that've come up to this point.
    Yup. Keep more digits in intermediate values or do the whole thing symbolically and only plug in values at the end.

    [Oops! I see that @kuruman got there ahead of me!]
     
  5. Feb 21, 2017 #4
    Thanks for the replies guys.

    Gneill I see what you mean with [itex] (2\Delta x /t) = V_x [/itex]

    not sure why I wrote that, it wasn't like that on the paper I was doing my work on, it was indeed

    [itex] \Delta x / t = V_x [/itex] because [itex] V_{0x} + V_x = 2V_x [/itex]

    So my method is good, great! One more question before I mark as "answered," so its best to do significant figures at the end the question, and keep all the intermediate values until then?
     
  6. Feb 21, 2017 #5

    gneill

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    Yes. Rounding should be done only for "presentation" values. In practice, keep two or three guard extra digits of precision in any intermediate values that you need to "store" on paper rather than keeping them in a calculator memory at full precision. Of course, it's always best to do the majority of the work algebraically, only plugging in values once the equations have been simplified and the number of operations minimized.
     
  7. Feb 21, 2017 #6

    haruspex

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    Why?
     
  8. Feb 21, 2017 #7

    gneill

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    I meant that if intermediate values could not be kept on the calculator for any reason then any values kept on paper should be be recorded with extra digits.
     
  9. Feb 21, 2017 #8

    haruspex

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    Ah, ok.
     
  10. Feb 21, 2017 #9
    I got 14.87 m/s using g = 9.8 m/s^2 and 15.13 m/s using g = 9.81 m/s^2
    Edit: had to correct units
     
  11. Feb 21, 2017 #10

    haruspex

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    I confirm those numbers. Interestingly, g=10 m/s2 gives 19.9 m/s, showing how sensitive the answer is to precision of the data.
    Going by significant digits, the least accurate input value is the cliff height. It could be anything from 55.5m, giving 17.5m/s, to 56.5m, giving 12.5m/s.
    This sensitivity arises because most of the 3.6s is taken by the descent of the rock. A small change in the latter produces a much larger percentage change in the time left for the sound to return.
     
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