Calculating the Depth of a Well and Velocity of a Rock Using Equations of Motion

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To calculate the depth of the well and the velocity of the rock, two time intervals must be considered: the fall time of the rock and the travel time of the sound. The equations of motion can be applied, using the known acceleration due to gravity and the speed of sound in air. The total time from the drop to hearing the splash is 20 seconds, which includes both intervals. Participants in the discussion emphasize the need to separate these times to solve for the well's depth and the rock's final velocity accurately. Clarifying the definitions of the time variables is crucial for the calculations.
Miri McPherson
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1. A rock is dropped into a well of an unknown depth. 20 seconds after the rock is dropped the splash is heard at the top of the well. What is the depth of the well? What is the velocity of the rock when it reached the bottom?2. X= Xo + Vo(t) + 0.5(a)(t^2)
V= Vo + a(t)
(V^2)= (Vo^2) + 2a(X-Xo)
Velocity of sound in air=343 m/s
Acceleration of gravity= -9.80 m/(s^2)

3. 0= Xo + 0.5(-9.80)(20-t of sound)
X= 343(t)
I feel like there isn't enough information, but I'm not sure. Any help would be greatly appreciated.
 
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Which time are you indicating with "t?" Fall time? Or, travel time of the splash noise? Or, total time?
 
Bystander said:
Which time are you indicating with "t?" Fall time? Or, travel time of the splash noise? Or, total time?
The time is from the time that the rock was dropped until the sound was heard.
 
Miri McPherson said:
The time is from the time that the rock was dropped until the sound was heard.

Welcome to the PF.

Divide it up into two times and two equations. Can you take a cut at doing that? :smile:
 

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