Dropping a Rock Down a Well to Find Depth - Brain Waves

[@ChrisEdelweiss]

Thank you! I actually needed this to figure out the depth of a hole in a D&D session to keep it as accurate as possible XD

[@danoftheroom381]

Same😂

[@ChrisEdelweiss]

@danoftheroom381 WOOO!! ＼(≧▽≦)／

[@josephhenderson403]

You can avoid the 3/2 power on the D by moving the fraction D/C to the left and then squaring. This gives you a quadratic which I found easier to deal with.

[@CAPTAINBUCKETS99]

You could also just use a measuring stick and avoid the math.

[@liolp808]

@@CAPTAINBUCKETS99 Don't think you went unnoticed! :)

[@harshavardhanshitole8591]

Hint: Formula to solve quadratic equation ax²+bx+c=0 is x=(-b±√(b²-4ac))/(2a)

[@Girlmossy]

My mans making physics videos next to a jet engine

[@boibotbot9030]

Could you please show the work that you did to solve for the distance with the given time? I didn't like that you just showed the answer without showing the process or at least the math process to solve for d. It made it even harder to solve the problem and until you can give an explanation how how you got that simple solution out of that horrendous equation you set up, I cannot recommend this video to anyone.

[@johndavid4007]

You need to learn how to solve these problems with a calculator. You need to now how to solve this by hand, but it it's just messy arithmetic and isn't really going to teach you anything you don't already know. Plus, you get messy cross powers of D (you'll get D^2 and D^1.5 on the left). That's why you should just graph it with D being along the x-axis, and you'll see that the resulting graph crosses the x-axis at 40.720261. That's your answer.

[@asian_buddie1405]

To be fair I was calculating the distance and I always got 44.145

[@saramarkovic9041]

​@@asian_buddie1405i also got 44.145 because i used this formula: D=ut+1/2×g×(t)² and u=0

[@immjayant]

Convert the equation to 0.0836D² - (58.86D/343) + 88.29 - 2D = 0 Now you have a quadratic equation solve it and you'll get two answers 40.71 meters and 159.53 meters

[@douglove2412]

Help! What is the easier method you teach the Search and Rescue people? I’m showing Realtors how to judge the depth of water with an ice cube drop. If we have the original Well Report, we know the total depth of the well, so we can then estimate how much water is standing in the well. But I’m suspicious of our math. Can you share yours, please?

[@johnnypayne9614]

A falling rock travels at 32.18 ft per second, if it hits the bottom of the well at 4 seconds then the well is 64 feet and 3.6 inches deep. If distant/depth is known, Divide the falling distance by 16. For example, if the object will fall 128 feet, divide 128 by 16 to get 8. Calculate the square root of the Step 2 result to find the time it takes the object to fall in seconds. Since you are trying to estimate water level and you know the depth of the well I will give you an easier method. Take some twine or rope that is 10 feet longer that the reported depth (in case the report is off) tie a weight at the end. Slowly drop the weight down the well. Once it hit bottom gently, pull the rope back up. Mark the water line on the rope then measure from the bottom of the weight.

[@specialdost2005]

Thanks for the detail working. However, i have a simple formula to calculate depth of anything. Drop the rock and note the time in seconds. (i.e 3 seconds) Then multiply the time with the same number. ( 3 x 3 = 9) Then multiply the answer with 16 then you will get the approx. depth in feet. (9 x 16 = 144 feet) If you convert 144 feet into meters, you will get approx 43.89 meters. That's it.

[@hadrielvalentino]

Hi Muhammad! May I know where this simple yet effective mode of calculation was derived from?

[@johndavid4007]

Except you've failed to account for the time spent by the sound waves traveling back up to the surface of the well. If the problem was as simple as you presented it here, it would be a a matter of plugging in two numbers into a basic equation, instead of doing the more sophisticated set-up that is necessary here.

[@johnnypayne9614]

@@johndavid4007 I think that why he is multiplying 3 seconds by itself (3*3=9) to account for the sound to return. I could be wrong but that is the only way this formula would give an approximate distance but not an exact distance. The video is good if you are working in a class room but not in a real work search and rescue where often time the team is not carrying a laptop computer or scientific calculator. Most of the time one has to do the calculations mentally as time is of the essence because the who fell person's life depends on it. This is also why most carry a 500 foot and 1,000 foot rescue ropes as most falls occured between those distances.

[@imposter709]

Why do you have to multiply the 9 with 16?? Can someone tell me that?

[@pradipkyada]

How could we find D from the equation...??? give us the solution ....

[@Ernest2]

How can I get the answer from the calculator?

[@cassandra1143]

this is so helpful😭

[@barzg4mingg448]

Is this what they call splashing method used to measure water level in wells?

[@marlonflores3962]

Please show how you solved for d

[@gingerfeest]

Wow! I didn't not that the speed of sound had that much of an impact on the final answer! Almost an 8% difference!

[@johndavid4007]

No idea how you're doing your math here, but the time spent by the traveling sound waves is 40.72/343, or 0.119 seconds so the fraction of the total time spent by the traveling sound waves is 0.119seconds /3 seconds = 3.97%, not 8%.

[@gingerfeest]

​@@johndavid4007 We are solving for D. If you assume the speed of sound has a negligible impact (i.e. c->infinity), then you would get D=44.1, which is ~8% more than the answer we get if we let c=343.

[@kaeez]

@@gingerfeestWrong in many ways. 1) [BLUNDER] When neglecting a mere value, you don’t change it to infinity (that is not neglecting but doing the opposite). 2) Even if you take c=0, that would break the equation as division by zero is undefined. You can’t change denominator values to 0.

[@gingerfeest]

​@@kaeezAssuming speed of sound has a negligible impact is perfectly consistent with c going to infinity. I'm not neglecting that sound has a speed, I'm saying that the higher the speed of sound is the less it will factor into the equation as a contribution towards the time it takes for you to hear something. Said another way, if instead of being asked how long it takes to hear the object hit the bottom, you were asked how long it takes to see the object hit the bottom, I would hope you wouldn't bother using the speed of light in your calculation, as its contribution is miniscule. And since it is minuscule, you naturally drop it, which is equivalent to setting c = infinity.

[@kaeez]

Well, you can't compare the speed of sound with the speed of light lol. It's less than 0.0001% of it. All the more, speed of sound in dry air is around 343 m/s approx., that's a pretty finite and comprehensible distance and we can certainly be involved in assessing displacements higher than that, so in that regards, it will have an impact whether that be fractional, it definitely adds up in the end if you're dealing with longer distances, let's say 2x 3x the mag. of speed of sound then that error would be in seconds if you're neglecting it. In the case of the scenario, discussed in the video, it won't be significant as the depth is well under the speed of sound but still have a slight fractional error but given that in real life we have to account for drag, this isn't yet another big deal. Plus, I don't understand mathematically how setting a value to infinity is neglecting it. In my math knowledge and experience, if I neglected a value in a physical equation, I would set it to zero thus dropping it from the equation. If you set it to infinity, it would break most equations. How is such a finite speed like the speed of sound be estimated to be infinity? Especially the kinematics one we're dealing with here which has velocity in the denominator. @@gingerfeest

[@SirSwed]

I had this question on a physics exam. I got 40.73 meters. Close enough, I hope.

[@mr.redneck2715]

I would recommend using an ice cube, usually there is wires ! you did not calculate bouncing off wires and the side of the casing.

[@30pranaypawar17]

@Julia Milford but the 7 seconds should be total time taken to reach down and sound to rise up. How do you know the time taken to reach bottom ?

[@30pranaypawar17]

@Julia Milford um hey. i was doing the same problem of stone dropped in well , can time taken by sound to return be taken negligible if splashing sound was heard at 7.7 seconds after the stone was dropped ???

[@30pranaypawar17]

@Julia Milford oh ok thanks 🤗

[@udaydangar4034]

How to solve this if the initial velocity of the rock is 2 m/s

[@aqmi]

Thank you so much

[@Dos-equis-guy]

So if you drop a feather in a well that will make a sound when it hits the bottom? And it takes 2 minutes for the feather to hit the bottom. I guess its a REALLLLLY deep hole. Does terminal velocity really not play a factor in this equation? You made it seem like this is a world with no air resistance.

[@Crosseyedhero]

The video did assume no air resistance. For shorter distances air resistance doesn't impact the calculation that much and can usually be ignored. If you did want to account for air resistance, the problem becomes much more painful to solve.

[@Dos-equis-guy]

@@Crosseyedhero good observation

[@Dos-equis-guy]

@@Crosseyedhero the air resistance slowed the sound waves as well. I didn't account for that in the past when I watched the video

[@Dos-equis-guy]

@@Crosseyedhero I think I was drunk when I was thinking about terminal velocity but I was thinking about something very interesting.

[@Dos-equis-guy]

I think I remember now. I was wondering how terminal velocity can be measured with something with very high air resistance, as it would be slowed very much along its descent. But if you are only using the sound it makes from the impact at the bottom of the well to to measure depth...doesn't that make the concept of measurement incorrect?

[@eonreeves4324]

what if you cannot see the rock land, you can only hear the sound?

[@fonsiakmac9796]

You have two eautions for height: h=g*(time of rock)^2/2 and h = (velocity of sound) * (time of sound). Now total time T = timeOfRock+timeOfSound, and we know T. Now you can calculate timeOfSound = height/velOfSound and you can subsititute it to previous equation and get timeOfRock = T-(height/(velOfSound)). And now you substitute time of rock to the first equation for h and get a polynomial equatino for height: h=(g*(T-height/timeOfSound)^2)/2.

[@mcimpoeru]

What was the background noise? It made the video difficult to follow....

[@johnnypayne9614]

This is horrible!!! We teach our search and rescue people a fair easier method because they don't always have calculators. There was no explanation here, no showing how to work the problem mentally as most of the time you are in the field and don't always have computers. Get some real work experience first.

[@flightlesslord2688]

Well said, what is said easier way?

[@johnnypayne9614]

@@flightlesslord2688 average speed of falling rock is 32.18 ft per second. If a rock falls down the well and hits bottom at 4 seconds then the depth of the well is 64 feet and 3.6 inches deep. Another way if distant is known but time isn't then Divide the falling distance by 16. For example, if the object will fall 128 feet, divide 128 by 16 to get 8. Calculate the square root of the Step 2 result to find the time it takes the object to fall in seconds.

[@Thryce23]

This is a physics homework problem, not the instructor’s problem you don’t comprehend it…

[@aHotdogThatComments]

With your formulas I get D=40.69293m, not 40,71m

[@safffff1000]

I'd use ice

[@rpaleg]

Forgets to calculate change in density of air as height/temp. changes, thus altering buoyancy of rock and speed of sound, also the air resistance against the rock. Smh

[@johndavid4007]

The density differential of air and change in speed of sound over a distance of 40.7 is negligible.

[@rpaleg]

@@johndavid4007 that's the joke

[@SephyCallum]

Guys I keep getting 3.1287026239 as my answer So I did D = 0.5*9.81*(3*3) = 44.145 Tfall = (Square root of 2*44.145)/9.81 = 3 Tsound = 44.145/343 = 0.1287026239 3 = Ttotal = 3 + 0.1287026239 = 3.1287026239m Where have I gone wrong in the equation I don’t get you got 40.71m 🤦🏻‍♀️

[@johndavid4007]

In your first equation ("D ="), you've plugged in 3 seconds for the FALLING time. But 3 is the TOTAL time spending falling AND the sound waves traveling back up the well. You don't know the amount of time spent just in free fall.

[@alucarddracula4285]

@@johndavid4007 so what is there amount of time spent in free fall?

[@matejkukl2290]

yea same

[@pretyinpunk1]

I watched the whole video and got nothing from it a simple answer would have been great

[@phil18751]

Hope this helps....

[@CINENACHII]

final answer should've been 44.06m

[@johndavid4007]

Nah. Check your work. The answer is approximately 40.720261 meters.

[@zakmatew]

@@johndavid4007 can you tell me if for the expression (3) we need to solve for both D’s?? For some reason I am not clear on that.

[@johndavid4007]

@@zakmatew D appears twice in the equation, but there aren't two Ds. Both Ds stand for the same variable (the depth of the well). It's just an equation in which D appears more than once, but both stand for the same thing.

[@zakmatew]

@@johndavid4007 thank you for the clarification!

[@kaeez]

@@johndavid4007But solving it algebraically would mean that D would have two values since it’s a second degree polynomial expression. You have to neglect one.

[@phantomtech287]

Super bro

[@ark2380]

Can anyone help with 28 seconds, I'm getting 12000 feet