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

  1. Aug 31, 2008 #1
    1. The problem statement, all variables and given/known data
    A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 4.7 s later. If the speed of sound is 340 m/s, how high is the cliff?


    2. Relevant equations
    d=vit+.5at^2
    vf^2=vi^2 +2ad
    vf=vi+at

    3. The attempt at a solution
    so, i don't know how to have both the speed of sound and the rock's acceleration counted. i don't know, please help!
     
    Last edited by a moderator: May 8, 2015
  2. jcsd
  3. Aug 31, 2008 #2
    well, break this up. you have two seperate ideas.
    how long does it take for the rock to hit the bottom?
    (rock accelerates down at g=-9.8m/s^2).
    add that to how long it takes for the sound to get back to you.
    (sound travels at constant velocity v=340m/s).
     
  4. Aug 31, 2008 #3

    Defennder

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    Write an equation for the time taken for the rock to hit the ocean using your kinematics equations. Then write another equation for the time taken for sound to travel upwards from the ocean to the top of the cliff. You can assume the speed of sound is constant in the latter. Name the time variable in the former equation t1 and t2 for the latter equation. You now have 2 simultaneous equations. Solve them and plug them into the first one to get h.
     
  5. Aug 31, 2008 #4
    . Edited .
    As other people said it.
     
  6. Aug 31, 2008 #5
    well, how do i know how long it takes?? i just need someone to start this. i have the initial velocity (0), the acceleration (9.81 m/s/s)..
    and equations vf=vi + at (but i'm missing vf and t)
    vf^2 = vi^2 + 2ad (missing d and vf)
    d=vi*t + .5*a*t^2 (missing d and t)
    so how do i figure it out??
     
  7. Aug 31, 2008 #6

    Defennder

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    You don't have to use your first equation. Use another one instead.
     
  8. Aug 31, 2008 #7
    ok but how can i use any of those when i'm missing 2 variables?
     
  9. Aug 31, 2008 #8

    Defennder

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    Sorry, I didn't mean to imply to use another one of your equations. What equation can you come up with relating the speed of sound to the distance traveled back up and the time taken?
     
  10. Aug 31, 2008 #9
    i have no idea how to make equations, i'm not that advanced in physics i think!
    i guess, you could say that '340 m/s * time' + 'vi*t + .5*a*t^2' is the total distance, but..idk. i just need a jumpstart.
     
  11. Aug 31, 2008 #10

    Defennder

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    Yes that's right. You need a separate variable for time taken for sound to travel back up, and one for the rock to drop to the waters.
     
  12. Aug 31, 2008 #11
    i don't know how to do this. i just need help. i know the answer already, i just want to see how to do it. could you show me? if you get the answer to be like 95ish, then it's right.
     
  13. Aug 31, 2008 #12

    Defennder

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    You have vst2 = h
    and h = 1/2 g t1^2 and t1 + t2 = 4.7 , by your equations. Express either t1 or t2 in terms of the other, substitute into the equations and solve for them and plug them into the equation to find h.
     
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