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Finding the depth of a well after dropping a rock

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data
    You drop a rock(from rest) into a well. You hear the rock hit the bottom 3.15 seconds after you dropped it. How deep is the well? Do not ignore the time for sound to travel back up to your ear.

    Velocity of sound = 343 m/s (constant), g = 9.8 m/s^2

    2. Relevant equations
    The equations I have to use are:
    1. V = Vo + a(t)
    2. deltaX = Vo + 1/2(a)(t^2)
    3. V^2 = Vo^2 + 2(a)(deltaX)
    4. deltaX = 1/2(Vo + V)(t)


    3. The attempt at a solution

    I tried using equation 4 and got: 1/2(0+343)(3.15) = 540.225m

    I have no idea if this is right or not. The entire portion involving sound is throwing me off. Where am I actually suppose to use the speed of the sound at?
     
    Last edited: Sep 27, 2012
  2. jcsd
  3. Sep 27, 2012 #2
    You drop the rock and it travels x distance to the bottom. When it hits the sound travels the same x distance to the top again.

    You could equate the travel down to the travel up and solve for 2 unknowns - the travel time down and up.
    Hope that's enough of a hint for you.
     
  4. Sep 27, 2012 #3

    Simon Bridge

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    Put the depth of the well as D.
    * put T1 = time it takes the rock to fall distance D... write out that equation (don't put any numbers in it yet.)

    * put T2 = time it takes for sound to travel back distance D ... write out the equation for that (again - don't put any numbers in)

    The time between dropping the rock and hearing the sound is T=T1+T2

    This should give you an equation where D is the only unknown - make D the subject and put the numbers in.
     
  5. Sep 27, 2012 #4
    I'm not sure I completely follow. Do you mean do something like:
    0(t1) + 1/2(-9.8)(t1^2) = 343(t2) + 1/2(-9.8)(t2^2)

    Where each of the equations should equal the same displacement? So then I should solve for t1 and t2?
     
  6. Sep 27, 2012 #5
    0(t1) + 1/2(-9.8)(t1^2) = 343(t2) + 1/2(-9.8)(t2^2)
    Can you guess what the implication of that part is?
     
  7. Sep 27, 2012 #6
    Hmm...Nothing really comes to mind from it other than it looks similar to 1/2(-9.8)(t1^2) on the other side of the equation. Sorry if I'm being a bit slow on this.
     
  8. Sep 27, 2012 #7
    That right part describes the sound portion of your little conundrum. And that bolded part implies that the sound is affected by gravity in the same way your rock would be.
    With sound being a wave, not a particle(or a collection thereof), as you might imagine that would not be true.
     
  9. Sep 27, 2012 #8
    So what you're saying is that the equation should actually look like: 1/2(-9.8)(t1^2) = 343(t2) since sound won't be affected by gravity?
     
  10. Sep 27, 2012 #9
    You drop a rock(from rest) into a well. You hear the rock hit the bottom 3.15 seconds after you dropped it. How deep is the well? Do not ignore the time for sound to travel back up to your ear.

    Velocity of sound = 343 m/s (constant), g = 9.8 m/s^2
    ------------------------------------------

    Known - T=t1 + t2=3.15s, vsound=343 m/s, g = 9.8 m/s^2

    Unknown- y depth of well, t1(top to down time), t2 (bottom to top time)

    Even though we do not know t1 and t2 but we know t1 + t2=3.15s

    You find the time for the rock to reach the bottom t1 and you add to time for the sound to travel(t2) to top =3.15s
     
    Last edited: Sep 27, 2012
  11. Sep 27, 2012 #10
    That is where I am getting tripped up at though. Since I only know the initial velocity and the acceleration for the rock, all of my formulas have 2 unknowns in them, which means I can't solve for time.
     
  12. Sep 27, 2012 #11
    You already have all the info you need, just not all in one place at the same time. So let me make it easier for you:
    2 unknowns, 2 equations.
     
  13. Sep 27, 2012 #12
    sorry deleting my post
     
  14. Sep 27, 2012 #13
    I'm still having a hard time visualizing how those two equations are helpful. Am I suppose to combine the two together somehow (maybe by replacing t1 and t2 with just T?)?
     
  15. Sep 27, 2012 #14

    Simon Bridge

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    It may help if Martix's quotes used the same notation:
    (Not his fault - they come from different places in the discussion.)

    (-g/2)(T12) = c(T2) ...eq.(1)
    T=T1+T2 ... eq.(2)

    c = speed of sound in air: you know this.
    g = acceleration of gravity: you know this.
    T = time from dropping stone to hearing sound: you know this.
    You don't know T1 and T2 ... solve the equations to find them.

    You do know how to solve simultanious equations don't you?

    The trouble is, if we just tell you the next step that would be the same as doing your homework for you.
     
  16. Sep 27, 2012 #15
    I solved for T1 = 3.15 - T2 and T2 = 3.15 - T1. I then substituted them into the first formula and solved for each.

    Which gave answers of:

    T1=3.31 sec and T2=-0.156 sec

    Which confuses me a bit. How can I have a negative value for T2?
     
  17. Sep 27, 2012 #16
    Well shame on us for not spotting it sooner and quoting blindly.
    No reason for that minus sign in front of "g"(no real difference in how you count the distance, be it top to bottom or otherwise).
     
  18. Sep 27, 2012 #17
    I do it this way to solve the problem.

    Question -How deep is the well?
    Let the depth of the well is y

    y=1/2 a t12=> t1=√2y/a .....(1)
    y=ct2 => t2=y/c.......(2)

    we know that t1 + t2=3.15s

    (1) + (2)
    t1 + t2=√2y/a + y/c =3.15s

    Add: only depth needed without computing for time.
     
    Last edited: Sep 27, 2012
  19. Sep 27, 2012 #18
    Alright, I resolved and got T1=3.01973 sec and T2=-0.130 sec

    Then I did:

    deltaX = 4.9(3.01973)^2 = 44.7m

    Is 44.7m actually how deep the well is?
     
  20. Sep 28, 2012 #19

    Simon Bridge

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    Just means that eq.(1) was not derived properly.
    I'll commonly go with whatever derivation the questioner comes up with on the grounds that I am not the one doing the calculation. As we've seen, it usually comes out in the end and encourages thinking about what the answers mean ;)

    The derivation back at post #8 was for displacement (a vector) treating "upwards" as a positive direction - otherwise a negative distance makes no sense either.

    Since the sound is coming back the other way, it's displacement is also going to be the opposite and we have to write ##\vec{y}_{sound}+\vec{y}_{stone}=0##.

    ##\vec{y}_{stone}=- \frac{1}{2}gT_1^2\hat{\jmath}##
    ##\vec{y}_{sound}= cT_2\hat{\jmath}##
    ##\vec{y}_{sound}+\vec{y}_{stone} = cT_2\hat{\jmath}- \frac{1}{2}gT_1^2\hat{\jmath} = 0 \Rightarrow cT_2 = \frac{1}{2}gT_1^2##

    Notice that the magnitudes of the two displacements are the same (same well, travelling in different directions) So we could just write

    ##D = |\vec{y}_{sound}| = |\vec{y_{stone}}|##

    It is a lot easier to think about this in terms of distances.

    Looking back to post #2 - I believe I warned about putting the numbers in too early.
    When you do that you can get confused and you can miss stuff... like that you don't need to compute the actual times ##T_1## and ##T_2## ;)
     
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