How High is the Cliff if a Rock Hits the Ocean in 3.2 Seconds?

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SUMMARY

The problem involves calculating the height of a cliff from which a rock is thrown, given that the sound of the impact reaches the observer in 3.2 seconds. The speed of sound is provided as 340 m/s. To solve this, three equations are necessary: one for the time it takes the rock to fall (t), one for the time it takes sound to travel back up (T), and a third equation relating the two times. This approach allows for the determination of the cliff's height using the principles of kinematics and sound propagation.

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Thanks everyone for helping with my last question. Here's another one...(I'm actually in a Physics class, not just asking for "s and g." My TA for the class doesn't have a clue what she's talking about.)

A rock is thrown off the edge of a sea cliff. 3.2 seconds later you hear the sound of the rock hitting the ocean. If the speed of sound is 340 m/s, how high is the cliff.

This question is baffling me because for the life of me I can't escape the feeling that there is not enough information to solve the problem. I don't need the answer; if someone could just help me with which equation to use to solve it.

Thanks a bunch...

Jason
 
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You can solve this using 3 equations.

One that relates the time, t , it took the rock to hit the ocean floor a distance,d, down the cliff. Assuming no initial velocity.



Another equation that relates the time, T , it took the sound to travel back up the distance, D, back up the cliff at a speed, V. This is assuming that it is traveling straight up vertically and ignoring the horizontal component of the sound's motion to the observer or person listening.


And one that relates the two times t and T giving you 3 equations with 3 unknowns.
 
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