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Sound Problem

  1. Jan 6, 2005 #1
    "On a day when the air temperature was 18oC a stone was dropped into the water from the top of a cliff. The splash it made when hitting the water below was heard 4.0 s later. How high is the cliff?"

    To me, it sounds like the 4.0 s includes the fall from the cliff, [seemingly] making this a lot harder than the problems I've done before. I don't really know what to do beyond getting the speed of sound.

    [tex]v = 332 + 0.60T_c[/tex]
    [tex]= 332 + 0.60(18)[/tex]
    [tex] = 343 m/s[/tex]
  2. jcsd
  3. Jan 6, 2005 #2
    If this problem is significantly more difficult than others you're doing, it MAY be asking you to find the height of the cliff if the sound was heard 4 seconds after the splash occurred. In that case you've completed all of the calculations by calculating the speed of sound, so good job!

    If that's not the case the calculations are more difficult. What do you think? Should you attempt to find the answer for a sound heard 4 seconds after you DROP the rock?
  4. Jan 6, 2005 #3
    Use convservation of energy to find the stone's speed upon impact with the water:
    [tex]mgh = \frac{1}{2}mv^2[/tex]
    Then use the usual motion equation to find how long it took the stone to hit the water from the moment it was released:
    [tex]v_f = v_i + at_1[/tex]
    t1 will be a function of h. You can also express t2, the time it took the sound to travel, as a function of h. You know the sum of t1 and t2, and eventually you will be able to find h. :)
  5. Jan 6, 2005 #4
    It's a level 1 class, and we're normally given problems beyond where we are, so I'm pretty sure it's 4 seconds after the rock is dropped and not after it hits the water.


    "Use convservation of energy to find the stone's speed upon impact with the water:"

    I'm not sure how I'm supposed to find the actual speed when I'm missing the height too... pretty sure I'm missing something...

    I just end up with something like [itex]19.6h = v^2[/itex].

    The rest of your post makes it sound like this was intented... I guess my math just isn't up to par.
  6. Jan 6, 2005 #5
    You're doing fine... I didn't mean you have to find the absolute value of the speed, what I meant (and what you did) was to find an expression for V that depends on h. Then use that expression to find the time from this equation:

    [tex]v_f = v_i + at_1[/tex]

    Eventually you should get:

    [tex]t_1 = \sqrt{\frac{2h}{g}}[/tex]

    Now, to find the time the sound wave travelled you use this equation:

    [tex]h = v_st_2[/tex]

    Lastly, you know that:

    [tex]t_1 + t_2 = \sqrt{\frac{2h}{g}} + \frac{h}{v_s} = 4[/tex]

    And now you can solve for h. :smile:
  7. Jan 6, 2005 #6
    BTW, I get an answer of about 68.5 meters, although my calculations could be wrong...
  8. Jan 6, 2005 #7
    Ah, cool. That makes sense. Thanks for you help!
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