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Homework Help: Looking for height of cliff

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data

    A ball is dropped from a sea cliff. 3.2 seconds later it is heard striking the water. The speed of sound is 340 m/s. How high is the cliff?

    I can't seem to get this question, every attempt seems to point to insufficient amount of information.
  2. jcsd
  3. Sep 25, 2011 #2


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    Explain what you've tried so far and why you believe there to be insufficient information.
  4. Sep 25, 2011 #3
    Well I've understood that the 3.2 seconds is not the time that the balls takes to reach the water. What we know for the ball is that the initial velocity is 0 m/s and acceleration is 9.8 (m/s)/s downwards. And since we assume no air friction = no air so the speed of sound from the bottom of the cliff is 340 m/s and it remains constant.

    delta t (total) = delta t( ball) + delta t(sound)

    what I get stuck on is that no matter how I try to get the time duration for any of these two constituents, there is not enough information. I have also tried to substitute the common denominator between these two, the delta d (distance). But then the physics makes no sense, and it completely becomes math, and still do not get the correct answer.
  5. Sep 25, 2011 #4


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    This is a good start. Seems to me there are two additional equations you can write (1) the physics relating the distance the ball falls and the time it takes, and (2) the time it takes sound to travel the same distance. You'll end up with three equations and three unknowns that can be readily solved by substitution. Hint: involves solving a quadratic equation. Do you know what the answer is suppose to be?
    Last edited: Sep 25, 2011
  6. Sep 25, 2011 #5
    Well I make the cliff height just under 50 metres high.

    Taking g as 10m/s2 h = 47 metres.
  7. Sep 25, 2011 #6
    I'm always getting around 50m and I don't know why.

    delta t (total) = [itex]\sqrt{2h/g}[/itex] + h/340

    3.2 = [itex]\sqrt{2h/9.8}[/itex] + h/340

    (3.2)2 = 2h/9.8 + h2/3402

    Is there something wrong here?
  8. Sep 25, 2011 #7


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    Yep, (a + b)2 doesn't equal a2 + b2. The pesky square root makes this approach too ugly. What I did was write two different expressions for h, equate them, then use the result to solve for t1 and t2. The value for h can then be calculated.
    Last edited: Sep 25, 2011
  9. Sep 25, 2011 #8
    MY GOSH. I can't believe I made such a simple mistake, thanks hotvette, I got the answer now.
  10. Sep 26, 2011 #9
    Solving the quadratic is not so difficult.

    Here is how to get rid of that pesky square root.

    3.2 = \frac{h}{{340}} + \sqrt {\frac{{2h}}{g}} \\
    {\left( {3.2 - \frac{h}{{340}}} \right)^2} = \frac{{2h}}{g} \\
    {\left( {\frac{{1088 - h}}{{340}}} \right)^2} = \frac{{2h}}{{9.81}} \\
    {\left( {1088 - h} \right)^2} = \frac{{231200}}{{9.81}}h \\
    {h^2} - 25744h + 1183744 = 0 \\
    h = \frac{{25744 \pm \sqrt {662753536 - 4734976} }}{2} \\
    h = \frac{1}{2}\left( {25744 \pm 25652} \right) \\

    [tex]{\rm{h = 46m}}\quad {\rm{ or}}\quad {\rm{ 256968m}}[/tex]
  11. Sep 26, 2011 #10
    This is the approach I used. Hotvette, would you mind showing yours?
  12. Sep 26, 2011 #11


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    Yours was more straightforward than mine.

    3.2 = t1 + t2

    h = 1/2 g t12 = 340t2

    Two equations in two unknowns. Solve for either t1 or t2, then calculate h.
    Last edited: Sep 26, 2011
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