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Simple Problem with Gravity and Time

  1. Mar 30, 2014 #1
    1. The problem statement, all variables and given/known data

    "A rock is dropped off a cliff into the water below. The sound of the splash is heard 3.0 s later. If the speed of sound is 332 m/s, calculate the height of the cliff above the water. (Note: the total time it takes for the rock to fall and the sound to travel upwards is 3.0 s)"

    Therefore,
    v1 = 0
    g = 9.8 m/s2
    Δt = 3.0 s
    vsound = 332 m/s

    2. Relevant equations
    FOR SOUND
    Δd = vsound * Δt2, where Δt2 is the time it takes from the sound to reach the top of the cliff from the bottom.
    FOR ROCK
    Δd = v1 * Δt + 0.5 * g * (Δt)2
    *the following equations may be useful but i doubt it*
    v2 = v1 + g * Δt
    (v2)2 = (v1)2 + 2 * g * Δd

    *assume no air resistance

    3. The attempt at a solution
    Δd = ?
    No clue.

    I figured that the answer should be 41 m, I believe, by trial and error but I would like to know how this can be solved in a normal way.
     
  2. jcsd
  3. Mar 30, 2014 #2

    Simon Bridge

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    The Δt for the rock is different from the Δt you set for the total time.
    Call it Δt1 ?

    Then Δt1+Δt2=what?

    Now you have three equations and three unknowns.
     
  4. Mar 30, 2014 #3
    It should be time for rock + time for sound = 3.0s and that the sound is not affected by gravity but rock is accelerating from 0 , at 9.8m/s2.
     
  5. Mar 30, 2014 #4

    Simon Bridge

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    Well done - now you can solve it.
    hint: all three equations have to be true simultaneously.
     
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