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

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To determine the height of the cliff from which a rock is thrown, the problem involves calculating the time it takes for the rock to hit the water and the time for the sound to travel back up. The total time of 3.2 seconds includes both the fall time of the rock and the ascent time of the sound. The speed of sound is given as 340 m/s, which is crucial for calculating the ascent time. By applying three relevant equations—one for the rock's fall, one for the sound's ascent, and one relating the two times—it's possible to solve for the cliff's height. Understanding these equations is essential for solving the physics problem accurately.
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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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