Kinematics, rock falling off cliff

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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 time must be divided into two parts: the time it takes for the rock to fall (t1) and the time for the sound to travel back up (t2). The speed of sound is given as 340 m/s, allowing for calculations of the distance the sound travels. The equations indicate that the distance fallen by the rock (D1) equals the distance the sound travels (D2). To solve for the height of the cliff, the relationship between t1 and t2 must be established, leading to the correct height calculation.
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so I am kind of stumped on this one here... A rock is dropped from a sea cliff and the sound of it striking the ocrean is heard 3.4 seconds later. If the speed of sound is 340m/s, how high is the cliff?

- now i got some variables, v1=0 m/s
time(total)=3.4s
a= -9.80m/s^2
delta d= ?

i got thinking that the total time it takes for the rock to fall then the sound to come back at you would be 3.4 seconds in total, so i don't know how to separate the time it takes the rock to fall from the time it takes the sound to actually get back to you.

by using the 3.4 seconds, i tried to get a distance which was off, at 113 m, but then realized that i need to find out how long it takes for the rock alone to fall, without the sound. i can't find a reasonable way to get this number.

please help!?
 
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Forget the rock. Just focus on the sound. You have a time it takes for the sound to travel a distance, and the velocity of that sound. Now take a look at your formula for speed.
 
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so I attempted this by saying 3.4s=t(rock falls)+t(sound back), then solved and got 2 answers, which were 112.2 m and -11908 m, and I tried seeing if the 112.2 m made any sense when I plugged It in, but it would take the rock more than 3.4 s to reach the bottom, so 112.2 m is impossible...any suggestions on the right equation to use?:s
 
You can completely ignore the rock. You have the speed of sound, and the time it takes the noise to travel the height of the cliff.

Your basic speed formula:
S=\frac{D_{Final}-D_{Initial}}{\Delta Time}
 
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but it's saying that the timer starts when the rock falls off the cliff of an unknown distance then sound travels back to u..in 3.4s, so would u use a formula 3.4=t(rock falls)+t(sound)?:)
 
Oh, sorry. I misread the question...and I have no clue how to keep going.
 
You started off right; 3.4s is the amount if time it takes the rock to fall from the cliff to the ocean (t_1[/tex]) <i>plus</i> the time it takes for the sound of the rock hitting the ocean to be heard from the top of the cliff (t_2[/tex]).&lt;br /&gt; &lt;br /&gt; What is the formula for the distance the rock falls (use t_1[/tex])?&amp;lt;br /&amp;gt; Call this D_1[/tex]&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; What is the formula for the distance the sound of the rock hitting the ocean travels (use t_2[/tex])?&amp;amp;amp;lt;br /&amp;amp;amp;gt; Call this D_2[/tex]&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; &amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; &amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; We know 2 things:&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; 1) t_1 + t_2 = 3.4 sec[/tex]&amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;gt; 2) D_1 = D_2[/tex]
 
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