Physics Problem

[jimfet]

Jump in the hole with the rock in your hand. Open your hand and the rock will be weightless before your eyes. Have a 50 ft. extension cord tied around your waste. Because it will get hung on something on the way down. This will save your life.

I carry one with me when I fly.

 

[Brian]

We had the math course in 10th grade and the application to a falling object in physics. Maybe the advantage of a public school education in Portsmouth NH back in the '60s.

 

[steerpike2]

An interesting aspect is that a quadratic equation will always give two results. Part of getting the right answer (if this were in an exam) would be to show both numeric results and pick the right one, i.e., that one that makes practical sense.

For this problem, the equation yields t1=10.44sec and t2=1.56sec
but also t1=-731,9sec and t2=743.9sec.

Using the second value of t2 for the sound wave gives a depth of 255.14km. So.. what kind of impractical situation can we think up where the hole is 255km deep and the rock hits the bottom 731 seconds before it was dropped?

 

[313guy]

An interesting aspect is that a quadratic equation will always give two results. Part of getting the right answer (if this were in an exam) would be to show both numeric results and pick the right one, i.e., that one that makes practical sense.

For this problem, the equation yields t1=10.44sec and t2=1.56sec
but also t1=-731,9sec and t2=743.9sec.

Using the second value of t2 for the sound wave gives a depth of 255.14km. So.. what kind of impractical situation can we think up where the hole is 255km deep and the rock hits the bottom 731 seconds before it was dropped?

That negative time value would also steer me away from that solution.:yes:

 

[toxcrusadr]

If you pick that answer you create a rip in the fabric of space-time, which allows the entire Vegan army to come pouring out of the hole on their multiple insectoid legs.

 

[bowtie427ss]

I haven't seen any mention of the rock's surface area to mass ratio. Surface area to mass ratio is noted in Newton's 2nd law.

In air(an atmosphere) this is quite relevant, different rocks will achieve different terminal velocities in an atmosphere.

So, i would ask the OP "what kind of rock was it?"

 

[toxcrusadr]

Um, it looked like hard rock to me, or maybe punk.

 

[quneur]

This problem can't be solved symbolically. There are too many variables. Note:
Tt = Td + Tu

where:
Tt = time total
Td = time downward - rock
Tu = time upwards - sound

Without knowing the ratio between Td & Tu there isn't a specific answer. Everyone is assuming 1/2at^2: the t being 12 second which it is not. But trying to solve for the first term:

Td = sqrt (2 * Dist / 1125)

where: Dist = distance, using 1/2at^2

for the second term, using Dist = Rate * Time:

Tu = Dist / 32

So first equation becomes: Tt = sqrt (2Dist/1125) + Dist/32

There no solution to the above equation (ADist^1/2 + BDist + C).

 

[aabottom]

This problem can't be solved symbolically. There are too many variables. Note:
Tt = Td + Tu

where:
Tt = time total
Td = time downward - rock
Tu = time upwards - sound

Without knowing the ratio between Td & Tu there isn't a specific answer. Everyone is assuming 1/2at^2: the t being 12 second which it is not. But trying to solve for the first term:

Td = sqrt (2 * Dist / 1125)

where: Dist = distance, using 1/2at^2

for the second term, using Dist = Rate * Time:

Tu = Dist / 32

So first equation becomes: Tt = sqrt (2Dist/1125) + Dist/32

There no solution to the above equation (ADist^1/2 + BDist + C).

The solution to the OP's orginal problem is analytic and has a closed form solution. I suppose I'll have to type up my derivation to the solution and post it here.

 

[steerpike2]

This problem can't be solved symbolically. There are too many variables.

Yes it can; there are not too many variables. You just end up with two simultaneous equations - nothing particularly difficult there.

Without knowing the ratio between Td & Tu there isn't a specific answer. Everyone is assuming 1/2at^2: the t being 12 second which it is not.

No such assumption was made, the 't' in 1/2at^2 is Td, not = 12
Td + Tu = 12 sec.

 

[steerpike2]

If you pick that answer you create a rip in the fabric of space-time, which allows the entire Vegan army to come pouring out of the hole on their multiple insectoid legs.

The rock hits the bottom before it was dropped... interesting twist since nothing in Newtonian mechanics and the maths puts any restriction on the direction of time travel.

 

[313guy]

The rock hits the bottom before it was dropped... interesting twist since nothing in Newtonian mechanics and the maths puts any restriction on the direction of time travel.

I had some stocks hit rock bottom before I dropped them.:yes:
Still hear the sound, too.

 

[toxcrusadr]

I had some stocks hit rock bottom before I dropped them.:yes:
Still hear the sound, too.

:lmao::lmao::lmao:

 

[quneur]

The solution to the OP's orginal problem is analytic and has a closed form solution. I suppose I'll have to type up my derivation to the solution and post it here.

Don't bother. It was late when I saw this and tried to attempt this. I came up with:

D^2/1024 - 3383/4800 * D + 144 = 0

Don't want to input into the quadratic formula.

 

[HeezDeadJim]

You don't need stinking math. Go get a couple rolls of twine, tie a weight to the end and unroll it down the hole

Somehow I find that hard to believe that you can put a little washer on the end of twine and send it down a "12-sec" hole. Do you realize how long of twine you'd have to have. And how high tensile/thick it would need to be to not snap on itself by the sheer weight of it's own length?!

Try dragging that spool of twine to the site.

 

[steerpike2]

Somehow I find that hard to believe that you can put a little washer on the end of twine and send it down a "12-sec" hole. Do you realize how long of twine you'd have to have.

Didn't we calculate that already? 535 metres. Since we are on Audio-karma, we can relate to that, since it is almost exactly one half of a 10.5-inch reel of 35 micron (LP) recording tape.

 

[HeezDeadJim]

Didn't we calculate that already? 535 metres. Since we are on Audio-karma, we can relate to that, since it is almost exactly one half of a 10.5-inch reel of 35 micron (LP) recording tape.

I realize that twine/weight suggestion was a joke. The spool of twine would be ridiculously big. But Even if you use two 10.5-inch reels of tape (these are much thinner than twine)....those aren't unwound. I don't think the tape would hold the metal reel if you let it hang that far down with gravity compounding itself. I've never tried this, though, so I could be wrong...who wants to try it out?

 

[usedto]

Another thing to consider:

It's a fairly universally accepted fact that the magnetic poles are shifting, and may someday switch ends as they have done before. During that phenomenon, the laws that affect gravity have a tendency to fluxuate. If so, the standards used for your calculation might be erroneous.

Of course, you could always ask the dude standing next to the hole with a shovel how deep it was.

 

[HeezDeadJim]

Another thing to consider:

It's a fairly universally accepted fact that the magnetic poles are shifting, and may someday switch ends as they have done before.

Wait...so YOU'RE telling me Santa used to live on the South Pole at one time?!?!? That's blasphemy, sir!

 

[sheltie dave]

You guys are SO antiquated you probably still use analog turntables as a music source.:sigh:

I would grab a surveyor's laser out of the tool box, center it over the hole, and take the average of three calibrated shoots.:sigh: Then I would power up my DeWalt toolbox CD boombox and see how many of the craft personnel still remember how to spell quadratic, let alone use the equation for which it is named. :lmao: