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Finding Velocity of a rock thrown into a well

  1. Dec 15, 2008 #1
    1. The problem statement, all variables and given/known data

    A rock is thrown downward into a well that is 9.17 m deep. If the splash is heard 1.15 seconds later, what was the initial speed of the rock? Assume the positive y-axis to point upwards

    2. Relevant equations
    Time of sound(Ts)= Distance/(343m/s)
    Time of stone(Tf)=Total time - Time of sound


    Vo=(Height Final - Height Initial)/Tf - 1/2 x g x Tgiven

    3. The attempt at a solution

    1.So basically I started off by finding the time it takes for the sound of the stone hitting the water to reach my ears (Ts):

    So i took the depth of my well divided by the speed of sound through air:
    9.17m/(343m/s)=.026735s

    2.Next i subtracted .026735 from the total time to find out how long it takes the rock to hit the water at the bottom of the well (Tf):

    1.15s-.026735s = 1.12327s

    3. Next using this new time (Tf) i try to figure out velocity, and this is where i think my math is getting screwed up, because i can't get the right answer. Here is my process:

    Using the formula Vo=(Height Final - Height Initial)/Tf - 1/2 x g x Tgiven

    Vo=(0-9.17m)/1.12327s - (.5)(-9.81m/s^2)(1.15s)


    The answer i get is -2.523m/s, which is not right. I am fairly certain the velocity should be negative, since the rock is traveling in the negative direction according to the Y axis parameters given, but other than that i don't know what other mistakes i might be making.

    I think my formula for step 3 might be slightly off, but i'm not sure where i'm making my mistake. Big thanks in advance for any help anybody can offer!!
     
  2. jcsd
  3. Dec 15, 2008 #2

    rl.bhat

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    Using the formula Vo=(Height Final - Height Initial)/Tf - 1/2 x g x Tgiven

    In this formula Tgiven should be Tf
     
  4. Dec 15, 2008 #3

    LowlyPion

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    Welcome to PF.

    Looks OK through 2.

    At 3. though you should consider that y = V*t - 1/2g*t2

    You know t from 2. and you know y = -9.17 so ...
     
  5. Dec 15, 2008 #4
    Thank you both very much for the help, i found the right answer. Problem solved!
     
  6. Jan 15, 2009 #5
    Hi, I am also having a problem with this question. It would be helpful if you could actually show the answer that you got (the right answer) and maybe what you were doing wrong intially. It would help me maybe to work through it! I did see that you were missing a 2 on the second t (t2) in your equation 3. Also, instead of using tgiven for equation 3, I am using the tfinal. For your question, I am calculating an intial velocity of -2.66m/s. Was that the answer?

    I guess we're using the same TB, but it just has different numbers for this question. My well is 9.0m deep, and the time until you hear the echo is 1.20sec.

    My first and second step are identical to yours: I get 0.02624sec for the echo, which leaves 1.2 - 0.02624 = 1.17376sec left for the rock to fall. I know that part is right, but then I suppose the third part is wrong somewhere.

    This is how I rearranged my formula for step 3. Is this right?:

    d = Vt + 0.5at2
    Vt = d - 0.5at2
    V = [d - 0.5at2] / t
    V = [(-9) - (0.5)(-9.8)(1.173762)] / 1.17376
    V = -1.92 m/s

    I know that the answer should be negative (it's been thrown downwards), yet that is the wrong answer. I think the math is right and the formula is rearranged just how I have rearranged every other formula, so I'm not sure where I'm going wrong.

    Thanks to anyone who can help!
     
    Last edited: Jan 15, 2009
  7. Jan 16, 2009 #6

    LowlyPion

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    Welcome to PF.

    Yes. -2.66 would be the answer for the original problem on my calculator.

    And with my rounding I get -1.91 for the math of your problem. One possibility might be that your problem answer doesn't expect you to account for the speed of sound which would certainly affect the answer enough to give an inexact result. (However, if they give you the speed of sound in air as part of the problem then they of course are indicating that you use that.)

    The error in the OP was that he was using 2 different time values in the equation - echo corrected and not.
     
  8. Jan 16, 2009 #7
    Wow, okay, thank you. Yes -1.91 is the answer and not -1.92 :surprised. I can't believe the computer wouldn't accept my first answer! I'll talk to my teacher and maybe he can change my mark. Yeah, the time for the echo is a factor is this question, so maybe that's why this question has a smaller tolerance for rounding errors because even if you didn't use the sound you would get pretty close to the correct answer anyways.

    Thanks again.
     
  9. Jan 16, 2009 #8

    LowlyPion

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    Glad it worked out for you. I wasn't being all that careful, as I rounded to 1.734 for the calculation. A .01 difference seems a bit slim for school work.
     
  10. Jan 16, 2009 #9
    Yeah, definately!! I've e-mailed my teacher, so I'll see what he says. It is a computer software thing though too, so the computer's answer may have had slightly different numbers too depending on how many decimals they used for their time (ie, using 1.7 instead of 1.7376). We'll see!
     
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