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Rock falls from cliff-Initial Velocity?

  1. Dec 10, 2016 #1
    1. The problem statement, all variables and given/known data
    A footballer kicks the rock off the cliff, with a horizontal velocity. The cliff's height is 40.0m. The rock falls in a lake. The footballer hears the "splash" 3.00s later. Sound travels through air with a velocity of 343 m/s. What is the rock's initial velocity?

    a = -g = -9.80 m/s^2

    2. Relevant equations
    Yf = Yi + Viy*t + 1/2*a*t^2
    Xf = Xi + Vix*t

    3. The attempt at a solution

    I set the ground as Y = 0, and the footballer's position as X = 0.
    Left to Right is (+) and Upwards is (+) as well.

    Yf = Yi + Viy*t + 1/2*a*t^2 <=> -40.0 m = -4.90 m/s^2 * t^2 <=> t = 2.857 s

    The whole time that it takes for the footballer to kick the rock, for it to fall in the lake and for him to hear the sound, is 3.00s. The rock falls inside the lake in 2.857s. Therefore, the time that the sound takes to travel from the lake to him is t' = 0.143 s.

    After that, is where I get stuck. The book's answer is 9.81 m/s, but mine is different. At first, I took sound's velocity as is, going with D = Vs*t' = 343 m/s*0.143s = 49.049 m. And then, I just did D = Vi*t <=> 49.049 m = Vi*2.857s <=> Vi = 17,168 m/s.

    Then I thought about it, and figured that it's not specified whether sound moves just vertically, so I took Vsx = Vs*cos(45) = 242.53 m/s and did the same things, but I still don't get the book's result.

    I'm probably missing something, but I've been at it for half an hour now, and I'm just not seeing anything else. It's exercise 21, M4 from the 8th Edition of "Physics for Scientists and Engineers".

    EDIT: I also tried this:

    Sound: R = Vs*t' = 343 m/s * 0.143s = 49.049 m

    So, I've created a triangle, and with Pythagoras' Theorem: D = sqrt(49.049^2-40^2) = 28.386 m

    Rock: D = Vi*t <=> 28.386m = Vi*2.857s <=> Vi = 9.935 m/s

    It's just that I'm not sure how many decimals to keep. I know the laws and rules, but the book sometimes goes with a more round number, and other times it keeps 3 decimals or so.
     
    Last edited: Dec 10, 2016
  2. jcsd
  3. Dec 10, 2016 #2

    PeroK

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    This is the correct method. I haven't checked all your working, but it looks close enough to the answer.

    When you plug the numbers in immediately, you will generally maximise the rounding error. The only way to get a more accurate answer is to tackle the problem algebraically and plug the numbers in at the end.
     
  4. Dec 10, 2016 #3

    PeroK

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    PS The answer I get with ##g = 9.8 ms^{-2}## is ##v = 9.9 ms^{-1}## (to one decimal place).
     
  5. Dec 10, 2016 #4
    I'll try that, and go with the usual methods of rounding errors this time. Thans for the tip!

    Yeah, that's what I get as well, with slight changes to the 0.0X decimal depending on whether I round the numbers or not.
     
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