Low Level Physics Problem 1D Motion

Click For Summary

Homework Help Overview

The problem involves a rock being dropped from a cliff, with the total time until the sound of the splash is heard being 3.4 seconds. The speed of sound is given as 340 m/s, and the objective is to determine the height of the cliff.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to establish a relationship between the time it takes for the rock to fall and the time for the sound to travel back. There are suggestions to use equations of motion and to derive formulas based on the height of the cliff.

Discussion Status

Some participants have provided guidance on how to set up the equations needed to solve the problem, while others have raised questions about the assumptions being made, particularly regarding the interpretation of the total time given.

Contextual Notes

There is a noted confusion regarding the total time being the sum of the fall time and the sound travel time, which affects how the problem should be approached. Additionally, there is an inquiry about the position of the observer in relation to the splash.

Maskkkk
Messages
7
Reaction score
0
47.

A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 3.4s later. If the speed of sound is 340 m/s, how high is the cliff?

Could somebody map this one out for me, it isn't making any sense.
 
Physics news on Phys.org
Solve the equations.
 

Attachments

  • Picture.GIF
    Picture.GIF
    2.1 KB · Views: 555
First, do you know a formula for distance an object falls in t seconds?
Solve that for t1 (the time until the rock hits the water) as a function of d (the height of the cliff).

Second, you are given that the speed of sound is 340 m/s so you can calculate the time, t2 necessary for the sound to come back up a distance d.

Since t1+ t2= 3.4 seconds, you can add those two formulas, involving d, to get an equation. Solve the equation for d.
 
Umm...Another question. Where in the world are you standing to hear the splash?
 
Since the acceleration of the rock is uniform i.e under the affect of gravity, couldn't you use the equations of motion?

u=0, a = 9.81, t = 3.4

s=ut+0.5xat^2
s=0.5*9.81*3.4^2
s=56.7m

I know that's not the information he was given but would it work? The only assumption I've made is that the rock was dropped from rest.
 
The time given in this question is not the time the rock takes to reach the ocean but is the total time the rock reaches the ocean plus the time for the splash sound to travel back to the dropper. So i don't think it would work here.
 
Leong said:
The time given in this question is not the time the rock takes to reach the ocean but is the total time the rock reaches the ocean plus the time for the splash sound to travel back to the dropper. So i don't think it would work here.

Ah yes, you are right. I didn't think :blushing: :smile:
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
3K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
8K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K