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1) A cave explorer drops a stone from rest into a hole. The speed of sound is 343 m/s in air, and the sound of the stone striking the bottom is heard 1.33 s after the stone is dropped. How deep is the hole?

I did 343 / 1.33. Is this correct?

No. Dividing a speed by a time gives you an acceleration rate. You have to write an equation for the time it takes to hear the sound. That time consists of two terms: stone drop time and sound travel time.

So I would set my problem up as

(13.3 - X/343)^2 = x/1.76

The only thing I am not sure of is how to solve for the x.

let ts be the time for sound to travel up, and td the time to fall to the bottom.

td+ts-1.33seconds

depth= 1/2 g td^2=4.9(1.33ts)^2

also, depth= 343m/s*ts

set these two equal, solve for ts. Notice it is a quadratic. Once you get the ts, solve for deptth

iam trying to find out when a rock drops onto a well 17 seconds later it hits the water how far down is the water level from the top of the well

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