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Properties of sound

  1. Sep 16, 2012 #1
    Hello,

    I'm calculating a physics problem and I'm supposed to calculate the height of a building given that:

    rock is dropped from the building,
    the sound of the rock hitting the ground is heard 4.8s later,
    the speed of sound is 350m/s,

    and I don't know how do I calculate the time since the rock hit the ground until it reached my ears.

    Thank you!
     
  2. jcsd
  3. Sep 16, 2012 #2

    Simon Bridge

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    distance over speed.
     
  4. Sep 16, 2012 #3
    I don't have the distance.
     
  5. Sep 16, 2012 #4

    Simon Bridge

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    That's right - distance is what you are trying to find.
    You do, however, have the total time, the speed of sound, and the acceleration due to gravity.
     
  6. Sep 16, 2012 #5

    Drakkith

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    Is it 4.8 seconds from the time you drop the rock until you hear the sound? Or 4.8 seconds from the time the rock hits the ground until you hear the sound?
     
  7. Sep 16, 2012 #6

    Simon Bridge

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    I'm interpreting that at t=0, rock is dropped, t=T=4.8s hear the sound.
    There are four equations and four unknowns ... waiting for OP to realise ;)

    Rock accelerates at g=9.8m/s, falls a height h, and hits the ground in time T1 at speed v making a "crack" noise.

    From this we see that OP can work out:
    v in terms of g and T1 ...that will be eq(1)
    h in terms of v and T1 ...that will be eq(2)

    The noise returns the same distance at speed c=350m/s in time T2:
    height of the building by the speed of sound and T2 ... (3)

    The total time between drop and sound is:
    T=T1+T2 ...(4)

    Four equations and four unknowns.
     
  8. Sep 16, 2012 #7
    Thank you.

    I have : -1/2aT12 = c(4.8-T1)

    I still can't find the answer.
     
  9. Sep 16, 2012 #8

    Simon Bridge

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    Well - technically T1 is one of the variables you want to eliminate, you want to keep h because that's what you want to find. I think you may have misplaced a minus sign too.

    Please list your four equations... number them.
    You will probably instinctively combine the first two... that's fine, you then have three equations and three unknowns.

    However: notice that you know a and c, so you can get T1 from that equation. Substitute that number into all your equations where T1 appears and you now have three equations and three unknowns.

    But it is better to to use the first three equations to eliminate all the unknowns except for h. Have you solved simultaneous equations before?
     
    Last edited: Sep 16, 2012
  10. Sep 16, 2012 #9
    You give me wonderful ideas on how to start solving physics problems effectively, but I need so much more practice.

    This time I got: T2= 2h/a + h2/c2

    however, I still can't get it. It's always so close.

    I don't know what more to ask you, I think you've told me all that you could.

    Thank you!
     
  11. Sep 16, 2012 #10

    Simon Bridge

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    That does not look right to me - I need to see your working or I cannot help you :(

    I suspect you actually had $$T=\sqrt{\frac{2h}{g}}+\frac{h}{c}$$ and you squared both sides. Is that the case?
    If so then that was a mistake.

    But look at the form of your final equation ... you've seen it before.
    What do you call it when the variable is squared in a function: what sort of function is it?
    It has form ##Ax^2+Bx+C=0##
     
    Last edited: Sep 16, 2012
  12. Sep 18, 2012 #11
    I think I have it now.

    1/2at12=ct2

    and t1= t-t2

    Then solve quadratic equation with a=1; b= -71.4; c=23.04

    Thank you so much!
     
  13. Sep 18, 2012 #12

    Simon Bridge

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    No worries.
     
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