Rock off cliff - calculate height by using speed of sound

  • Thread starter Bikengine
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You are climbing in the High Sierra where you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of this cliff, you drop a rock from the top and 8.50s later hear the sound of it hitting the ground at the foot of the cliff.

Ignoring air resistance, how high is the cliff if the speed of sound is 330 m/s?

v = u + at
s = ut + 1/2at^2
d = st



I am confused because the rock takes 8.5 secs to fall, and I can't figure out how to determine at what point in those 8.5 seconds the rock hits the bottom, and the sound of it starts travelling up to the top of the cliff
 

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  • #2
CAF123
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I am confused because the rock takes 8.5 secs to fall, and I can't figure out how to determine at what point in those 8.5 seconds the rock hits the bottom, and the sound of it starts travelling up to the top of the cliff
The 8.5 s refers to the time that you hear the rock hit the bottom. So the time for the rock to hit the bottom, t1, is less than 8.5s and the remaining (8.5 - t1)s is the time for the sound of the rock hitting the bottom (or echo) to reach you.
 
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I have the answer now, thanks very much!
 

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