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If the speed of sound is 340m/s, how high is the cliff?

  1. Jan 19, 2005 #1
    Physics Help Needed! (work shown just need an extra push in the right direction! )

    Hi, I'm currently taking a Physics 100 level course, so this question should be no problem for you smart people out there. lol My grade 12 teacher definetly did not prep me for this semester - or maybe I just don't remember my physics. Either way I'm having lots of trouble and was hoping someone would give me a hand.

    Here's the problem: A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard 3.2s later. If the speed of sound is 340m/s, how high is the cliff?

    Well what I did is I used the following formula:
    d = 0.5(Vf + Vi)t
    d = 0.5(340m/s - Om/s)(3.2s)
    d = 544m

    but that seems too simple of a problem to me - not to mention the GINORMOUS height I got for the cliff. It just doesn't seem reasonable! help! :confused:
     
    Last edited: Jan 19, 2005
  2. jcsd
  3. Jan 19, 2005 #2
    please help!
     
  4. Jan 19, 2005 #3

    dextercioby

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    You can find the same problem in this forum.Can u solve a quadratic?

    It involves knowing the speed of sound in air...

    Daniel.
     
  5. Jan 19, 2005 #4
    The speed of sound is given. I need to know how to interpret the data given to me - - because obviously I've done something very wrong. :yuck:
     
  6. Jan 19, 2005 #5

    dextercioby

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    Oh,yes u did...Consider the fact that the total time (3.2s) is the sum between the time of the free fall and the time it takes for the sound to come back your ear...

    HINT:For the 2 times,the distance is the same and is unknown...

    Daniel.
     
  7. Jan 19, 2005 #6
    I honestly dont know - I know that you're trying to help me but I just don't understand - I'm not given the velocity of the falling rock - nor am i given the time that it takes for the rock to fall and hit the water - it doesn't seem like there is enough information to solve the problem.

    the only knowns I have are:

    a of rock = -9.8m/s^2
    v of sound = 340m/s
    total time = 3.2s
    d = ???
     
  8. Jan 19, 2005 #7

    learningphysics

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    Let t1 be the time it takes for the rock to drop to the bottom.
    Let t2 be the time it takes for the sound to travel back to the top.

    Can you come up with some equations using t1, t2 and d?
     
  9. Jan 19, 2005 #8
    so something like:
    t1 + t2 = 3.2, 3.2-t1=t2 (and vise-versa)

    and then use

    x = xo + (vo)(t1) + 0.5(a)(t2)^2
    x = 0 + 0 + 0.5(-9.8m/s^2)(3.2 - t2)^2
    x = -15.68 - (-4.9)t2

    and

    x = (Vf + Vi)t2
    x = (340m/s^2 + 0m/s^2)(3.2 - t1)
    x = 1088m/s - (340)t1

    like that? and then what?
    man this is getting frusterating!
     
  10. Jan 19, 2005 #9
    wow 57 views and only 7 posts! hehe please help me! I'm really stuck!
     
  11. Jan 19, 2005 #10
    Ooops\, sorry ignore this. :P
     
    Last edited by a moderator: Jan 19, 2005
  12. Jan 19, 2005 #11
    i'm really confused
     
    Last edited: Jan 19, 2005
  13. Jan 19, 2005 #12
    Hmm, I am not sure if this is right, but here's a try:

    [tex]
    t_1 + t_2 = 3.2
    [/tex]

    ...

    [tex]
    v = v_0 + gt_1
    [/tex]

    rearrange so that

    [tex]
    t_1 = \frac{v - v_0}{g}
    [/tex]

    using work energy theorem

    [tex]
    v = \sqrt{2gh}
    [/tex]

    and you also know that

    [tex]
    t_2 = \frac{h}{v_s}
    [/tex]

    So then you can solve for h.
     
  14. Jan 19, 2005 #13

    learningphysics

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    This formula should only have one time: t1.

    x= vo*t1 + 0.5*a*t1^2

    Since v0=0, this becomes

    x= 0.5*(9.8)t1^2. (Since I'm only dealing with distances here and since the rock is dropping in the downward direction I use +9.8 instead of -9.8)

    I don't know where you got this formula... did you mean
    x=0.5(Vf+Vi)t2 ? You can use that. Remember here where are talking about the distance travelled by the sound back upward, and the speed of sound is constant.
    so Vf=Vi=340.

    Or the easier way is to just use
    x= v*t2
    x=340*t2

    Ok. So there are 3 equations you've got:

    t1+t2=3.2
    x=0.5*9.8*t1^2
    x=340*t2

    Can you solve for x using these 3 equations? You've got 3 equations in 3 unknowns.
     
  15. Jan 19, 2005 #14
    wow! that should definetly help - I'll get back to you once I've tried again - thanks!
     
  16. Jan 19, 2005 #15
    okay using those three equations:

    i) t1 + t2 = 3.2s
    ii) x=0.5(9.8m/s^2)(t1)^2
    iii) x = 340m/s(t2)


    ii) x = (4.9m/s^2)(t1)^2

    iii) x = 340m/s(3.2s - t1)
    x = 1088m - 340m/s(t1)

    so now I need to somehow combine the two t1's and solve for x

    would that work? or am I way off track?
     
  17. Jan 19, 2005 #16
    also how would i combine the t1's ?
     
  18. Jan 19, 2005 #17
    it seems really complicated becuz in one equation the t1 is squared
     
  19. Jan 19, 2005 #18
    learningphysics are you still around?
     
  20. Jan 19, 2005 #19
    I'm sooo close to being done guys - please can someone explain how to combine the two t1 values so that I can solve for x - and thus solving this problem?
     
  21. Jan 19, 2005 #20

    learningphysics

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    I'd solve for t1 in equation iii (so get t1 in terms of x) and then plug it into equation ii. So you get one equation all in x.

    Yes, it is kind of complicated... but stick with it. Let me know what you get.

    EDIT: Much better to solve for t1 in equation iii, and then plug into equation ii. Makes things easier. No square root.

    Then solve the quadratic!
     
    Last edited: Jan 19, 2005
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