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If a rock is dropped off of a sea cliff

  • Thread starter kbrowne29
  • Start date
12
0
If a rock is dropped off of a sea cliff, and the sound of the rock hitting the water is heard 3.4 seconds later, how tall is the cliff, assuming the speed of sound is 340 m/s.
What I've been trying to do is break the problem up into 2 parts, one for the rock going down towards the sea, and one for the speed of sound going up the cliff. The problem is that I can't seem to substitute the right things into the equation x = x0 + v0t + .5at^2. For the going down part, the equation looks like the following:
X= 0 + 0 + (.5)(9.8)t^2. And for the going up part, I'm not sure whether the initial velocity of the speed of sound is 340 m/s, or whether it is zero. If the initial speed were 340 m/s, then the equation would (?) look like the following:
x = 0 + 340t + 0.
This is where I get stuck, and I don't know what to do with the two equations. I would greatly appreciate the help. Thanks.
 
Last edited by a moderator:

HallsofIvy

Science Advisor
Homework Helper
41,711
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Sound does not accelerate and is not affected by gravity!

Your calculation for the rock falling: X= 0 + 0 + (.5)(9.8)t^2 is correct. For the sound coming back up, X= 340t.

What you can do is write these as X= 0 + 0 + (.5)(9.8)t1^2,
X= 340 t2 and t1+ t2= 3.4.

You can solve the two equations (.5)(9.8)t1^2= 340 t2 and t1+ t2= 3.4 for t1 or t2 separately and then use the equations to find X.
 
12
0
Thank you very much for the help; your idea of creating the equation v1+ v2 = 3.4 is exactly what I needed to finish the problem off. I hadn't previously thought about solving the equations silmultaneously like that.
 

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