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Easy One Dimensional Kinematics

  1. Aug 22, 2009 #1
    1. The problem statement, all variables and given/known data

    A rcok is dropped from a sea cliff and the sound of it striking the ocean is heard 3.4 s later. If the speeds of sound is 340 s^-1 m how high is the cliff.

    2. Relevant equations

    X = Xo + Vo t + 2^-1 a t^2

    3. The attempt at a solution
    Ok 340 average velocity and sense there's no acceleration??? then it is also equal to the Vo right?

    So i had this

    X = Vo t

    plugged in the numbers and got

    1156 m which is wrong

    my book tells me the answer is 52 m why???

    I think my data might be wrong from how i read the problem

    I took Vo to be 340 s^-2 m

    t to be 3.4 s

    acceleration to be zero
     
  2. jcsd
  3. Aug 22, 2009 #2

    Doc Al

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    Staff: Mentor

    Think of this in two steps:
    (1) The rock falls and hits the ocean.
    (2) The sound from the splash travels from the ocean to the top of the cliff.
    Each takes time, adding up to the total time (which is given).
     
  4. Aug 22, 2009 #3
    ok thanks i should be able to do that
     
  5. Aug 22, 2009 #4
    ok when I solved for the time that it took for sound to travel

    I got

    62.41 s or .1852 s

    then when I calculated distance

    I got

    21219.4 m or 62.968 m
    and it's obviously the second one but will I know which one is the extraneous soultion because some times it might no be so obvious...

    dang looks like i did it wrong to =[
     
  6. Aug 22, 2009 #5

    Doc Al

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    Show what you did.
     
  7. Aug 22, 2009 #6
    ok well i was suppose to get 52 m

    does some one know...

    um I got .1852 s for the time it takes for the sound to reach the top
    i'm taking it that I went wrong there

    For my quadratic equation I got the following...

    a = 4.90 m/s^2

    b = -306.7 m/s

    c = 56.64 m

    can anyone give me a hand???

    y = ax^2 + bx + c
     
  8. Aug 22, 2009 #7
    I established...

    X = (9.80 m/s^2 (t1^2))/2

    x = 340 m/s (t2)

    were t1 is the time it took the rock to drop
    t2 is the time it took for the time to reach the top of the cliff

    t1 + t2 = 3.4 s

    t1 = 3.4s - t2

    (9.80 m/s^2 ((3.4s - t2)^2))/2 = 340 m/s (t2)
    (9.80 m/s^2 (11.56 s^2 + t2^2 - 2(3.4 s)t2))/2 = 340 m/s (t2)
    ((9.80 m/s^2 (11.56 s^2 + t2^2 - (6.8 s)t2))/2 = 340 m/s (t2))2
    19.6 m/s^2(23.12 s^2 + 2 t2^2 - (13.6 s)t2) = (680 m/s) t2
    453.2 m + (39.2 m/s^2)t2^2 - (266.6 m/s)t2 -(680 m/s) t2= (680 m/s) t2 - (680 m/s) t2
    453.2 m + (39.2 m/s^2)t2^2 - (266.6 m/s)t2 -(680 m/s) t2 = 0
    453.2 m + (39.2 m/s^2)t2^2 - (266.6 m/s)t2 -(680 m/s) t2
    453.2 m + (39.2 m/s^2)t2^2 - (946.6 m/s)t2 = 0
    (39.2 m/s^2)t2^2 -(946.6 m/s)t2 + 453.2 m = 0

    t2 = (-b +/- sqrt(b^2 - 4ac))/(2a)

    t2 = (946.6 m/s +/- sqrt((-946.6 m/s)^2 - 4(39.2 m/s^2)453.2 m))/(2(39.2 m/s^2))
    t2 = (946.6 m/s +/- sqrt((-946.6 m/s)^2 - 4(39.2 m/s^2)453.2 m))/(2(39.2 m/s^2))
    t2 = (946.6 m/s +/- 907.9 m/s^2)/(2(39.2 m/s^2))
    t2 = (946.6 m/s +/- 907.9 m/s^2)/(78.4 m/s^2)

    t2 = 23.65 s
    t2 = .4936 s

    x = Vo t
    x = (340 m/s)23.65 s = 8041 m

    x = Vo t
    x = (340 m/s).4936 s = 167.8 m
     
    Last edited: Aug 22, 2009
  9. Aug 22, 2009 #8

    Doc Al

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    Recall the two steps I mentioned in my earlier post? You need to express the time for each step in terms of the unknown distance. Then solve for that distance.

    Let me know when you're done editing your post.
     
  10. Aug 22, 2009 #9
    oh I thought i did
     
  11. Aug 22, 2009 #10

    Doc Al

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    I only know what you tell me. Are you done editing your post?
     
  12. Aug 22, 2009 #11

    Doc Al

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    Looks good. Solve for t2. You'll get two solutions, only one of which is the one you want.
     
  13. Aug 22, 2009 #12
    I'm done got nonreal answer...
     
  14. Aug 22, 2009 #13

    Doc Al

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    My equation is the same as yours. It has two real solutions for t2, only one of which is physically meaningful for this situation.
     
  15. Aug 22, 2009 #14
    ok well I finsihed i put my final answers and they were a lot more than 52 m
    don't know were i went wrong either
     
  16. Aug 22, 2009 #15

    Doc Al

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    Show your solution to that quadratic. You must have made a mistake when solving it. (I got 52 m.)
     
  17. Aug 22, 2009 #16

    Doc Al

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    So far, so good.
    Where did that 2 come from on the right hand side?
    This (and what follows) is wrong.

    Note: Don't keep going back and editing earlier posts--that makes your thinking very hard to follow.
     
  18. Aug 22, 2009 #17
    (9.80 m/s^2 ((3.4s - t2)^2))/2 = 340 m/s (t2)
    (9.80 m/s^2 (11.56 s^2 + t2^2 - 2(3.4 s)t2))/2 = 340 m/s (t2)
    I multiplied the whole equation by the inverse of the inverse of the 2 to get rid of it
    ((9.80 m/s^2 (11.56 s^2 + t2^2 - (6.8 s)t2))/2 = 340 m/s (t2))2

    when I did that i got this
    19.6 m/s^2(23.12 s^2 + 2 t2^2 - (13.6 s)t2) = (680 m/s) t2
     
  19. Aug 22, 2009 #18

    Doc Al

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    But you just multiplied the right hand side by 2:
    The factor of 2 (in the denominator) still remains on the left hand side. Correct this error.
     
  20. Aug 22, 2009 #19
    Is this correct I'll edit it

    (9.80 m/s^2 ((3.4s - t2)^2))/2 = 340 m/s (t2)

    (9.80 m/s^2 (11.56 s^2 + t2^2 - 2(3.4 s)t2))/2 = 340 m/s (t2)

    ((9.80 m/s^2 (11.56 s^2 + t2^2 - (6.8 s)t2))/2 = (340 m/s)t2)2

    multiply the whole thing by 2 gave me this

    (19.6 m/s^2(23.12 s^2 + 2 t2^2 - (13.6 s)t2)) = (680 m/s)t2
     
  21. Aug 22, 2009 #20

    Doc Al

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    Still wrong. You have an extra factor of 2 on the RHS.

    If you wish to multiply by 2, you must multiply both sides by 2.
     
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