Calculating the height of a cliff given only time to the splash

[KiwiThief]

Hello everyone, I am a just now starting to study physics and I am quite the newbie, but I am loving it! I am not doing this for school, I study as a hobby in my spare time, but I imagine this is the right place to post this exercise anyway!

Here it is:
We have a rock dropped from a cliff that falls in the ocean, and after 3.2 seconds, we can hear the splash of the rock into the water. We are looking for the height of the cliff!

So, we have:
t = 3.2 s (total time)
g = 9.80 m/s²
v of sound = 340 m/s (or -340 m/s if we put it in the same equation as the rock, perhaps?)
v₀ of the rock = 0 m/s
Using the formula for constant acceleration, we can easily find the height of the cliff while ignoring the time it takes for the sound to travel from the water back to the person that threw the rock:
x = x₀ + v₀t + 1/2at²
(I am using x instead of y for convenience, even though the more commonly accepted one here would probably be y)
since x₀ = 0 and v₀ = 0, the equation ends up being

x = 1/2at²
and when we solve it we get the answer x = 50.176m

The obvious problem is that the time for the rock to fall is not t, it is actually t - t₂ (t₂ being the time it takes for the sound to return), and I am finding it hard to figure out an equation that would combine all the information I have (which is not much) in one and give me the correct answer. I was thinking about a lot of possible solutions, for example using the fact that the sound and the rock are both traveling the same distance (x₀ to x and then x to x₀), but no luck so far. Please help!

 

[phinds]

Try looking at it from the point of view of a variable that represents the total time.

Also, this should be using the homework template

Also, also, welcome to the forum

 

[jedishrfu]

Welcome to PF!

In future please use the homework template.

You have the right idea the distance the sound travels back to you is the same as the distance the stone has fallen.

So t1 is the time taken to fall and t2 is the total time taken to fall the entire distance and to return the splash sound.

t2 = (time_to_fall) + (time_to_hear_sound) = t1 + (t2-t1)

write a similar equation for 2x and solve for t2 using some algebra...

 

[phinds]

t2 = (time_to_fall) + (time_to_hear_sound) = t1 + (t2-t1)

My post was intended to lead him to this rather than just dump it in his lap.

 

[KiwiThief]

Thank you very much for the help and the warm welcome! I was closer to the answer than I imagined, which is an encouraging thought!
And of course I will make sure my future posts are in the proper format!

 

[jedishrfu]

My post was intended to lead him to this rather than just dump it in his lap.

I was working on my response and didnt see your reply until after i posted. I wanted to clarify the usage of time which he understood already but couldn't quite articulate. Perhaps i gave more help than was needed but i didnt think it was too much. Thanks for letting me know though.

 

[phinds]

I was working on my response and didnt see your reply until after i posted...

Yeah, I've had that happen to me. I saw that our posts were close and should have considered that we sometimes get sidetracked in the midst of responding and someone slips in before our post goes in and this kind of thing comes up.

 

[bobui]

Welcome to PF!

In future please use the homework template.

You have the right idea the distance the sound travels back to you is the same as the distance the stone has fallen.

So t1 is the time taken to fall and t2 is the total time taken to fall the entire distance and to return the splash sound.

t2 = (time_to_fall) + (time_to_hear_sound) = t1 + (t2-t1)

write a similar equation for 2x and solve for t2 using some algebra...

What exactly do you mean by '2x' to solve for, under what premise? and wouldn't t1 + (t2-t1) leave you with t2 again?
Could you please try to clarify on what you were trying to say? It kinda made me more confused than I was reading your reply

 

[fresh_42]

What exactly do you mean by '2x' to solve for, under what premise? and wouldn't t1 + (t2-t1) leave you with t2 again?
Could you please try to clarify on what you were trying to say? It kinda made me more confused than I was reading your reply

If you take a look at the date of this thread, you'll probably recognize, that the chances to receive an answer are close to zero. Therefore I'll lock this thread now. If you are really interested in an answer, please be so kind and open a new thread, preferably in this homework forum and by the use of our (automatically inserted) template. Especially section (3) of the template is important to us, but we'd like all of them to be used.