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

[jshaner858]

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

 

[quasi426]

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.

 

[quicknote]

Hello Jason,

I'm glad to see that you are taking an active role in your physics class. It's great that you are seeking help and trying to understand the concepts better.

In order to solve this problem, you will need to use the equation for vertical motion, also known as the equation of motion. This equation is:

y = y0 + v0t + 1/2at^2

Where:
y = final position
y0 = initial position (in this case, the height of the cliff)
v0 = initial velocity (in this case, the velocity at which the rock was thrown off the cliff)
a = acceleration (in this case, due to gravity)
t = time (in this case, 3.2 seconds)

You can rearrange this equation to solve for y0, the initial position or height of the cliff:

y0 = y - v0t - 1/2at^2

Now, let's plug in the given values:

y0 = 0 - 0 - 1/2(-9.8)(3.2)^2
y0 = 0 - 0 - (-50)
y0 = 50 meters

So, the height of the cliff is 50 meters. I hope this helps and good luck with your physics class! Keep asking questions and seeking help when needed.