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Height of cliff using energy and speed of sound

  1. Sep 28, 2014 #1
    1. The problem statement, all variables and given/known data

    A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 4.0s later.
    how high is the cliff
    2. Relevant equations
    .5mv^2+mgy = E
    v = v0 + at
    v = d/t

    3. The attempt at a solution

    The question says after the rock is DROPPED so i need to figure out the time it took to drop and the time it took for the sound to travel back and those two times added together will be 4 sec
    t1 is the time it takes for the rock to hit the water. using energy
    mgh = .5mv^2
    gh = .5v^2
    to find v
    v = v0 + at
    v = gt
    so
    gh = .5g^2t^2
    2h=gt^2
    2h/g = t^2
    t1= sqrt (2h/g)

    to find t2
    v = d/t
    v=h/t
    t2= h/v where v is speed of sound

    t1 + t2 = 4
    sqrt(2h/g) + h/v = 4
    sqrt(2h/g) = 4 - h/v
    2h/g = (4-h/v)^2
    2h/g = 16 - 8h/v + h^2/v^2
    2h/g +8h/v = 16 - h^2/v^2
    2hv/gv + 8hg/gv = 16 - h^2/v^2
    [(2hv+8hg)/gv] = 16 - h^2/v^2
    [(2v+8g)/gv]h = 16 - h^2/v^2
    h^2/v^2 + [(2v+8g)/gv]h - 16 = 0
    (1/v^2)h^2 +[(2v+8g)/gv]h -16 = 0
    (1/(343^2))h^2 + [(2*343+8*9.8)/(9.8*343)]h -16 = 0
    (8.50x10^-6)h^2 + .227 h -16 = 0
    using quadratic formula
    [-.227 +/- sqrt((-.227^2)-4(8.50x10^-6)(-16))]/(2* (8.50x10^-6)
    (-.227 +/- .228)/ (1.7*10^-5)

    -.227 + .228/ (1.7*10^-5) = 58.8m = h

    mastering physics wants it to 2 sig figs so i submitted 59 m but it was marked wrong. where did i make a mistake?
     
  2. jcsd
  3. Sep 28, 2014 #2

    haruspex

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    Have you tried substituting your answer back into the starting equations to calculate the two times?
     
  4. Sep 28, 2014 #3
    t1 = sqrt(2h/g) = sqrt((2*58.8)/9.8) = 3.46
    t2 = h/v = 58.8/343 = .171
    t1 + t2 = 4
    3.46 + .171 = 3.63
    so its not quite 4
     
  5. Sep 28, 2014 #4

    haruspex

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    Well short of 4. So try subbing it in the last quadratic equation. Narrow down the location of the error that way.
     
  6. Sep 29, 2014 #5
    ended up finding it by trial and error that way. still trying to figure out where i went wrong mathematically
     
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