How High is the Cliff if a Rock Splashes Water in 3 Seconds?

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To determine the height of the cliff, the total time of 3.0 seconds must be divided into two parts: the time it takes for the rock to fall (Δt1) and the time for the sound to travel back up (Δt2). The equations for the distance fallen by the rock and the distance traveled by sound are Δd = 0.5 * g * (Δt1)² and Δd = vsound * Δt2, respectively. By setting up the equation Δt1 + Δt2 = 3.0 seconds, and substituting Δt2 with (3.0 - Δt1), the height of the cliff can be calculated. The solution involves solving these equations simultaneously to find the correct height, which is estimated to be around 41 meters. The discussion emphasizes the importance of correctly partitioning the total time to solve the problem accurately.
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Homework Statement



"A rock is dropped off a cliff into the water below. The sound of the splash is heard 3.0 s later. If the speed of sound is 332 m/s, calculate the height of the cliff above the water. (Note: the total time it takes for the rock to fall and the sound to travel upwards is 3.0 s)"

Therefore,
v1 = 0
g = 9.8 m/s2
Δt = 3.0 s
vsound = 332 m/s

Homework Equations


FOR SOUND
Δd = vsound * Δt2, where Δt2 is the time it takes from the sound to reach the top of the cliff from the bottom.
FOR ROCK
Δd = v1 * Δt + 0.5 * g * (Δt)2
*the following equations may be useful but i doubt it*
v2 = v1 + g * Δt
(v2)2 = (v1)2 + 2 * g * Δd

*assume no air resistance

The Attempt at a Solution


Δd = ?
No clue.

I figured that the answer should be 41 m, I believe, by trial and error but I would like to know how this can be solved in a normal way.
 
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The Δt for the rock is different from the Δt you set for the total time.
Call it Δt1 ?

Then Δt1+Δt2=what?

Now you have three equations and three unknowns.
 
It should be time for rock + time for sound = 3.0s and that the sound is not affected by gravity but rock is accelerating from 0 , at 9.8m/s2.
 
Well done - now you can solve it.
hint: all three equations have to be true simultaneously.
 

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