A rock dropped from a cliff using speed of sound

[ndnkobra]

Homework Statement

A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 5.1 s later. If the speed of sound is 340 m/s, how high is the cliff?

Homework Equations

d=v/t(didnt work)

The Attempt at a Solution

My first attempt at this problem was to try that formula above, but it was not the correct way to do this problem. I then realized that the time i had was the total time, and that i needed to find the time it took for the rock to hit the water, or the time it took for the speed of sound to go up, which i need distance for. I am now stuck there :(

 

[Hootenanny]

Welcome to Physics Forums.

You are indeed correct that we first need to determine the time taken for the rock to fall. How do you suppose that we could work that out?

 

[Maroc]

From all i know is that you have to divide your distance by 2. Because you are calculating it from bottom to top.

 

[Hootenanny]

From all i know is that you have to divide your distance by 2. Because you are calculating it from bottom to top.

And why would we want to divide it by two?

 

[Chronos]

Not so simple, rock drops due to gravitational acceleration whereas sound travels at the speed of sound. Division by two is not the answer. Integrate and compare the time lag. 5.1 seconds sounds suspiciously close.

 

[fatra2]

Hi there,

You have to divided the problem in two parts. One where the rock is in free fall. Then, once the rock hits the ocean, the sound is emitted and travels at constant speed.

Cheers

 

[redargon]

This problem could be interpretted in two ways:

1) the rock is dropped and the total time elapsed is 5.1 seconds until the sound is heard. A two part problem where the two parts will have to be equated to each other in terms of distance.

2) the rock is dropped and the dropper sees it hitting the ocean and hears the sound 5.1s later. In this case it is simply the distance that a sound wave would travel in 5.1 seconds at 340m/s

 

[Hootenanny]

This problem could be interpretted in two ways:

1) the rock is dropped and the total time elapsed is 5.1 seconds until the sound is heard. A two part problem where the two parts will have to be equated to each other in terms of distance.

2) the rock is dropped and the dropper sees it hitting the ocean and hears the sound 5.1s later. In this case it is simply the distance that a sound wave would travel in 5.1 seconds at 340m/s

I would say that it's quite clear that the only correct interpretation is #1. The question clearly states that the rock is dropped and then 5.1 seconds later a sound is heard. It does not say that the rock is dropped it hits the ocean and 5.1 seconds later a sound is heard.

 

[Maroc]

And why would we want to divide it by two?

oh nvm got confused with another problem. Sorry.

 

[ndnkobra]

Okay, so i took the time to look over this problem and separated the problem into two parts. I let d be my distance, so d = (0)(t1) + 1/2(9.8)t1^2. t1 being my first time. I then took d = vt and used the speed of sound for this problem, so d = (340m/s)(t2). t2 being my second time. i then made the problems equal to each other so (340)t2 = 1/2(9.8)(t1^2). I seem to be stuck here, can anyone help me with the algebra in this problem?

 

[ndnkobra]

Getting warmer...t1+t2=5.1s, so t2=5.1-t1. I therefore can substitute in 340(t2) as 340(5.1-t1)=4.9(t1^2)