Free fall question -- Dropping a pebble into a well to figure out the depth

Click For Summary
The discussion focuses on calculating the depth of a well using the time it takes for a pebble to fall and the sound to travel back up. The initial calculations incorrectly combine the time for both stages, leading to an erroneous depth of 2620.1 meters. To solve the problem correctly, the time for the pebble to fall and the time for the sound to travel must be treated as separate variables. The total time of 7 seconds should be divided into these two components, requiring three equations to relate depth, fall time, and sound travel time. Properly addressing these stages is essential for accurate results.
rabsta00
Messages
3
Reaction score
0
Homework Statement
Hikers came across a deep narrow well and decided to find its depth. They dropped a pebble into the well and heard the sound of the pebble hitting the bottom of the well in 7s. Using this information they found the depth of the well. How deep is the well? Assume that the velocity of sound is 340m/s.
Relevant Equations
v=v0-gt
y=y0+v0t-0.5gt^2
t=7s
v=340m/s
v=v0-gt
340=v0-9.8x7
v0=408.6
d=d0+v0t-0.5gt^2
d=0+408.6x7-0.5x0.8x7^2
d=2620.1m

this is what I've done but apparently is incorrect, not sure why or what the proper method is
 
Physics news on Phys.org
rabsta00 said:
Homework Statement:: Hikers came across a deep narrow well and decided to find its depth. They dropped a pebble into the well and heard the sound of the pebble hitting the bottom of the well in 7s. Using this information they found the depth of the well. How deep is the well? Assume that the velocity of sound is 340m/s.
Relevant Equations:: v=v0-gt
y=y0+v0t-0.5gt^2

t=7s
v=340m/s
v=v0-gt
340=v0-9.8x7
v0=408.6
d=d0+v0t-0.5gt^2
d=0+408.6x7-0.5x0.8x7^2
d=2620.1m

this is what I've done but apparently is incorrect, not sure why or what the proper method is
You seem to have completely confused the two stages.
First, the pebble falls to the water. Write an equation for that involving the depth and the time taken.
Second, the sound travels to the top. Write an equation for that involving the depth and the time taken.
Make sure to use different symbols for different variables.
 
Welcome rabsta00! :cool:
Please, consider that everything described in the problem happens in 7 seconds.
The stone travels downwards with steadily increasing velocity, then the waves of sound travel upwards at constant velocity.
 
There are three unknowns: the depth of the well, the time it takes the pebble to drop and the time it takes for the sound to travel up the well. Since there are three unknowns, it will take three equations to solve the problem. The first equation relates the depth of the well to the time it takes the pebble to drop. The second relates relates the depth of the well to the time it takes for the sound to travel of the well. The third relates the two times.
 
Modify ##y=y_0+v_0t+½gt^2## to account for the time it takes for the sound to travel from the bottom of the well to the top. What is the implication of the wording "they dropped a pebble" in respect of this equation? (Take downward as + so that g is positive in the equation)
 
Last edited:
neilparker62 said:
Modify ##y=y_0+v_0t+½gt^2## to account for the time it takes for the sound to travel from the bottom of the well to the top. What is the implication of the wording "they dropped a pebble" in respect of this equation? (Take downward as + so that g is positive in the equation)
There's a lot more wrong than that. Note the equation that includes the speed of sound and g.(!)
 
If have close pipe system with water inside pressurized at P1= 200 000Pa absolute, density 1000kg/m3, wider pipe diameter=2cm, contraction pipe diameter=1.49cm, that is contraction area ratio A1/A2=1.8 a) If water is stationary(pump OFF) and if I drill a hole anywhere at pipe, water will leak out, because pressure(200kPa) inside is higher than atmospheric pressure (101 325Pa). b)If I turn on pump and water start flowing with with v1=10m/s in A1 wider section, from Bernoulli equation I...

Similar threads

Replies
3
Views
1K
Replies
2
Views
6K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 4 ·
Replies
4
Views
9K
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
19K
  • · Replies 4 ·
Replies
4
Views
19K
Replies
2
Views
4K