Calculate Time for Stone to Hit Bottom of 180m Well at 20C

[samisoccer9]

Homework Statement

1. A woman drops a stone into a well 180 m deep. If the air temperature is 20 oC, how much time will elapse between the moment the stone is dropped and the moment the sound of the stone hitting the bottom of the well is heard? [6.6s]

[/B]

Homework Equations

Vsound = 331.4 + .59 x T
vt = d or one of the big five equations regarding kinematics?

The Attempt at a Solution

i find the new velocity of sound to be 343.2 m/s using the above equation, and i then tried to multiply the distance, 180m, by two, as sound travels there and back, and then take the equation t = d/v, t = 360/343.2, but it didnt give the right answer. i believe it has do to with kinematics regarding freefall maybe, or just incorporating accel. answer is 6.6s ^^

 

[wabbit]

The sound doesn't travel there and back, unless the woman is shouting at the stone and listening for the echo. Instead, first the stone drops and hits the bottom (how long does this take ? ) then the sound travels to the surface (how long does that take ? )

 

[Isaac0427]

I apologize if this is incorrect, but I believe that you do not have enough information. It would be the time that it takes the rock to travel to the bottom, plus the time it would take the sound to go from the bottom to top of the well. So, you can get one side of that by having the speed of sound at about 343.4 m/s (I believe to be more exact it would be 343.59 but let's keep it at 343.4 for now), and set up a proportion 343.4m/1s=180/xs. This would make x= about .52 seconds. Now, you would need to calculate how long it would take for the stone to get to the bottom. We can use D=vt for this, but we don't have a value for v. Let's plug in ma for v (because F=ma, and this is gravity so F=s). Your problem does not state the mass of the stone or the acceleration of the stone.

 

[wabbit]

@Isaac0427, you are correct that the initial velocity of the stone would change the answer, but I think we should assume that "drops", with no other qualification, means with zero initial velocity.

The mass of the stone however is not relevant as far as its fall under the influence of gravity is concerned.

 

[Isaac0427]

@Isaac0427, you are correct that the initial velocity of the stone would change the answer, but I think we should assume that "drops", with no other qualification, means with zero initial velocity.

The mass of the stone however is not relevant as far as its fall under the influence of gravity is concerned.

Really? I thought that a larger object falls with more velocity than a larger one.

 

[samisoccer9]

Now earlier today I got to the same part as @Isaac0427, with 0.52s being the time for sound to travel. Then I was still stick at attempting the stone, as since t = D/V, i was not sure which velocity i could use to solve for this part of the problem

 

[Isaac0427]

Now earlier today I got to the same part as @Isaac0427, with 0.52s being the time for sound to travel. Then I was still stick at attempting the stone, as since t = D/V, i was not sure which velocity i could use to solve for this part of the problem

That's exactly my point. There is no velocity in this. For the answer to be 6.6 seconds the average velocity would have to be about 29.6 m/s.

 

[wabbit]

Really? I thought that a larger object falls with more velocity than a larger one.

No, this was the point of Galileo's experiment (see link in my previous reply). The effect of gravity is a force proportional to the mass of the stone, but then the effect of a force is an *acceleration* inversely proportional to that mass - and the two cancel out.

To go back to the original problem, the remaining issue is to determine the duration of the fll, starting from an initial velocity of 0m/s and applying a constant acceleration of 9.81 m/s^2.

 

[samisoccer9]

@wabbit How would the remainder of the solution be carried out then ?

 

[wabbit]

If you assume a constant acceleration $g=9.81m/s^2$ and starting velocity $v(0)=0m/s$, what is the velocity $v(t)$? Given that, what is the depth $x(t)$ reached at time t after the drop ?