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    Belleville's Avatar
    Belleville Posts: 1, Reputation: 1
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    #1

    Mar 21, 2007, 08:17 PM
    physics grade 11
    A cave explorer drops a pebble into a cavern to find how far down it is to the water. The stone is heard hitting the water 1.8 s after release. If the speed of sound in the air in the cave is 345 m/s, how deep is the cavern?

    The answer is 15 m. How do you get this number using d=4.9(t)^2 and d=v(t)?
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #2

    Mar 22, 2007, 01:13 AM
    well...

    put in 345 to d = vt so d = 345t, this is applicable to the part where the sound is going up.

    d = 4.9t^2 is applicable to the rock going down, the ts are different.

    so now we can add them:



    is that enough to help you solve it?
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