Who likes physics problems

[Dritnul]

Here's the problem i need to solve:

You drop a rock down a well 1.5 secs later u hear it splash. How deep is the well?

assume acceleration due to gravity is 9.8 m/s and speed of sound in air is 330m/s

 

[KLin]

42 fathoms?

 

[computeerrgghh]

eleventy billion parsecs

 

[TecHNooB]

11.025 meters! I ignored that sound mumbo jumbo

 

[invidia]

(1/2)*9.8m/s*(1.5)^2 = 11.025m ?


I don't understand how the speed of sound comes into play. Or are you trying to confuse us?

 

[Dritnul]

Originally posted by: invidia
(1/2)*9.8m/s*(1.5)^2 = 11.025m ?


I don't understand how the speed of sound comes into play. Or are you trying to confuse us?

the rock travels down and the sound travels back up its not a trick

 

[spidey07]

Is the rock on a conveyor belt that constantly matches it's acceleration due to gravity?

What is the air pressure and the medium for the speed of sound?

Air we taking into account the air pressure as the rock travels down the well?

 

[MrDudeMan]

Originally posted by: invidia
(1/2)*9.8m/s*(1.5)^2 = 11.025m ?


I don't understand how the speed of sound comes into play. Or are you trying to confuse us?

the time it takes for splash sound to travel up the well is relevant.

 

[Dritnul]

air resistance is ignored...
its as if u have a rock freefalling in a vacuum and then the sound traveling back up the well through a normal atmosphere with a velocity of 330m/s

 

[Yossarian]

t1 = time rock falls
t2 = time for sound to travel up the well
s = well depth

t1 + t2 = 1.5
s = .5(9.8) * t1^2
s = 330 * t2

3 equations, 3 unknowns

10.56m

btw if you're gonna be picky enough to include the speed of sound in the problem, you should state how far above the well your hand is when you drop the rock, and how tall you are

 

[Mark R]

The air resistance is likely to be more significant than the speed of sound.

 

[spidey07]

Originally posted by: Dritnul
air resistance is ignored...
its as if u have a rock freefalling in a vacuum and then the sound traveling back up the well through a normal atmosphere with a velocity of 330m/s

But it doesn't work like that.

no sound in a vacuum. Did you just magically make air appear? Are we taking into account gravity on this air, because it surely affects the speed of sound.

 

[Dritnul]

Originally posted by: spidey07

Originally posted by: Dritnul
air resistance is ignored...
its as if u have a rock free falling in a vacuum and then the sound traveling back up the well through a normal atmosphere with a velocity of 330m/s

But it doesn't work like that.

no sound in a vacuum. Did you just magically make air appear? Are we taking into account gravity on this air, because it surely affects the speed of sound.

its a simplified problem its not meant to be real world

 

[MrDudeMan]

Originally posted by: Mark R
The air resistance is likely to be more significant than the speed of sound.

yes but this is very typical general physics problem. in order for the calculations to work out without much more complex math, you have to assume the free-falling object experiences no drag, but also that the sound travels at a known speed through a known medium.

 

[Dritnul]

Originally posted by: MrDudeMan

Originally posted by: Mark R
The air resistance is likely to be more significant than the speed of sound.

yes but this is very typical general physics problem. in order for the calculations to work out without much more complex math, you have to assume the free-falling object experiences no drag, but also that the sound travels at a known speed through a known medium.

Thank you

 

[spidey07]

if I had to guess...24.3 meters.

 

[Dritnul]

Originally posted by: spidey07
if I had to guess...24.3 meters.

that isnt possible the rock wont fall that far in less than 1.5 seconds

 

[spidey07]

Originally posted by: Dritnul

Originally posted by: spidey07
if I had to guess...24.3 meters.

that isnt possible the rock wont fall that far in less than 1.5 seconds

Good. You're learning. Now work it out.

I guestimated. 1s = 9.8 M traveled. 2s = 9.8*2 = v of 20. two seconds = 30 meters traveled. At that distance speed of sound will not be a significant factor.

Just work it out.

 

[Dritnul]

Originally posted by: spidey07

Originally posted by: Dritnul

Originally posted by: spidey07
if I had to guess...24.3 meters.

that isnt possible the rock wont fall that far in less than 1.5 seconds

Good. You're learning. Now work it out.

I guestimated. 1s = 9.8 M traveled. 2s = 9.8*2 = v of 20. two seconds = 30 meters traveled. At that distance speed of sound will not be a significant factor.

Just work it out.

theres an equation that will work through trial/error it is 10.567m deep, but i was hoping for some1 who knew the equation

 

[Yossarian]

why don't you read through the thread again

 

[spidey07]

Originally posted by: Dritnul
theres an equation that will work through trial/error it is 10.567m deep, but i was hoping for some1 who knew the equation

You have two equations. Solve for time.

 

[invidia]

x/330 + xt = 1.5s
and
0.5(9.8)t^2 = x

Solve for x

 

[Mark R]

Basiclly, derive 2 expressions for the depth of the well - one based on the down leg, one on the up leg.

Reduce the 2 equations into a quadratic.

Solve.

Subsitute the positive root into the expressions for the depth of the well.

 

[Dritnul]

Originally posted by: invidia
x/330 + xt = 1.5s
and
0.5(9.8)t^2 = x

Solve for x

yea that looks right

 

[invidia]

Originally posted by: Dritnul

Originally posted by: invidia
x/330 + xt = 1.5s
and
0.5(9.8)t^2 = x

Solve for x

yea that looks right

no it's not, I just realize that x*t = meters seconds which isn't the right dimensions and won't add with seconds. Sorry.

I have it now. It's not two equations and just one quadratic equation. Do the work, I don't want to give the answer