Low Level Physics Problem 1D Motion

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A rock dropped from a cliff takes a total of 3.4 seconds for the sound of it hitting the ocean to be heard. The problem involves calculating the height of the cliff using the time it takes for the rock to fall and the time for the sound to travel back up. The speed of sound is given as 340 m/s, which allows for the calculation of the time it takes for sound to return. The total time is the sum of the fall time and the sound travel time, leading to the need for two equations involving the height of the cliff. The discussion emphasizes that the total time includes both the fall and sound travel, clarifying that the initial assumption of using uniform acceleration alone is insufficient.
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A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 3.4s later. If the speed of sound is 340 m/s, how high is the cliff?

Could somebody map this one out for me, it isn't making any sense.
 
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Solve the equations.
 

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First, do you know a formula for distance an object falls in t seconds?
Solve that for t1 (the time until the rock hits the water) as a function of d (the height of the cliff).

Second, you are given that the speed of sound is 340 m/s so you can calculate the time, t2 necessary for the sound to come back up a distance d.

Since t1+ t2= 3.4 seconds, you can add those two formulas, involving d, to get an equation. Solve the equation for d.
 
Umm...Another question. Where in the world are you standing to hear the splash?
 
Since the acceleration of the rock is uniform i.e under the affect of gravity, couldn't you use the equations of motion?

u=0, a = 9.81, t = 3.4

s=ut+0.5xat^2
s=0.5*9.81*3.4^2
s=56.7m

I know that's not the information he was given but would it work? The only assumption I've made is that the rock was dropped from rest.
 
The time given in this question is not the time the rock takes to reach the ocean but is the total time the rock reaches the ocean plus the time for the splash sound to travel back to the dropper. So i don't think it would work here.
 
Leong said:
The time given in this question is not the time the rock takes to reach the ocean but is the total time the rock reaches the ocean plus the time for the splash sound to travel back to the dropper. So i don't think it would work here.

Ah yes, you are right. I didn't think :blushing: :smile:
 
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