ChatterBank5 mins ago
Distance Measurement
I watched a movie the other day in which an explorer dropped vertically a heavily loaded rucksack into a cavern. When he heard it hitting the bottom, he announced that the cavern was "about 3/4 mile" deep to his colleagues.
Is it possible to measure depth in this way? Wouldn't the weight of the rucksack have some relevance?
Thank you.
Is it possible to measure depth in this way? Wouldn't the weight of the rucksack have some relevance?
Thank you.
Answers
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No best answer has yet been selected by bigbanana. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.An object accelerates as it falls until (because of air resistance) it reaches its terminal velocity; it then continues to fall at constant speed. The terrminal velocity for an object depends on a number of factors, including particularly weight, size and shape.
If you knew that the rucksack did not reach terminal velocity, then you could calculate the depth of the cavern. Time Team once attempted this for a well
http://books.google.c...0well%20depth&f=false
If the rucksack does reach terminal velocity, and you know the value of this, you can calculate the depth in two stages. First, calculate the distance fallen while accelerating as above; from this, calculate the time accelerating. Then calculate the distance fallen at constant speed for the remaining time.
This assumes a direct fall; if the rucksack often bounces of the sides of the cavern, it will be slowed down a bit, so results will be less certain.
If you knew that the rucksack did not reach terminal velocity, then you could calculate the depth of the cavern. Time Team once attempted this for a well
http://books.google.c...0well%20depth&f=false
If the rucksack does reach terminal velocity, and you know the value of this, you can calculate the depth in two stages. First, calculate the distance fallen while accelerating as above; from this, calculate the time accelerating. Then calculate the distance fallen at constant speed for the remaining time.
This assumes a direct fall; if the rucksack often bounces of the sides of the cavern, it will be slowed down a bit, so results will be less certain.
Galileo's is known for dropping two cannonballs of different masses from the Leaning Tower of Pisa and showing that, as far as the eye could tell, they fell at the same pace.
Thought experiements show this has to be. If you linked the 2 together would they fall at a 3rd new rate ? Or would the 2 balls fall at different rates and snap the connection ? No, it falls just the same speed/acceleration regardless of mass.
Thought experiements show this has to be. If you linked the 2 together would they fall at a 3rd new rate ? Or would the 2 balls fall at different rates and snap the connection ? No, it falls just the same speed/acceleration regardless of mass.
I think air resistance and wind would indeed slow down the ping pong ball. But if we are talking about something like a large brick and a similar sized (but much heavier) lead block, the two would fall at the same speed.
I have seen one of these calculations before for finding the depth of a well and I thought the calculation was flawed as it it didn't take account of the time taken for the sound of the splash/crash to travel back up the well- that time can be significant for a deep well given that sound travels at only 300 metres/second
I have seen one of these calculations before for finding the depth of a well and I thought the calculation was flawed as it it didn't take account of the time taken for the sound of the splash/crash to travel back up the well- that time can be significant for a deep well given that sound travels at only 300 metres/second
Wildwood: Since the acceleration of a body = Force/mass, then assuming the mass stays the same means that increasing the force increases the acceleration.
However in the specific case of gravity, the force is proportional to the mass, which means that the force due to gravity=constant x mass.
So to go back to the first formula:
Acceleration of a body under gravity=constant x mass/mass = constant.
So the FORCE on a more massive body IS larger than the force on a smaller body, but the ACCELERATION is the same.
This only applies to gravity. Think of the force produced by a car engine; in this case the same force on a more massive car will produce a smaller acceleration and vice versa.
However in the specific case of gravity, the force is proportional to the mass, which means that the force due to gravity=constant x mass.
So to go back to the first formula:
Acceleration of a body under gravity=constant x mass/mass = constant.
So the FORCE on a more massive body IS larger than the force on a smaller body, but the ACCELERATION is the same.
This only applies to gravity. Think of the force produced by a car engine; in this case the same force on a more massive car will produce a smaller acceleration and vice versa.
Thanks vascop. I am still not convinced that a 12" aluminium sphere would reach the ground at the same time as a lead 12" sphere, providing they are dropped from high enough for both to reach their terminal velocity. My thinking is that the alumium one would have a slower terminal velocity than the lead one so once both have reached maximum acceleration, the lead one would continue to fall faster than the aluminium one.
Never mind, it seems to be beyond me. Thanks for explaining.
Never mind, it seems to be beyond me. Thanks for explaining.
Thank you all for your very interesting comments. However, I'm still puzzled because if the depth of the cavern could be measured in this way, what is the formula for the calculation ie how did he provide an estimate of the depth given that he only seemed to mentally count the seconds until the rucksack hit the bottom with a thud?
Thanks again.
Thanks again.
//how did he provide an estimate of the depth given that he only seemed to mentally count the seconds until the rucksack hit the bottom with a thud?//
Experience.
All the same, I think I'd opt for a slightly longer rope. I wouldn't care to climb down a 3/4 mile long rope only to discover I'd underestimated the distance to the bottom by a measly 5%.
Experience.
All the same, I think I'd opt for a slightly longer rope. I wouldn't care to climb down a 3/4 mile long rope only to discover I'd underestimated the distance to the bottom by a measly 5%.
@bigbanana:
Ignoring air resistance (the guy's only interested in a rough value) then the distance in metres =0.5 x 9.8 x time (in seconds) x time(in seconds).
So if it takes 10 seconds to hear the thud, then the distance is
0.5 x 9.8 x10 x 10=490 metres.
As factor says, sound takes a second to travel 300metres so we should subtract about 1.5 seconds off this time giving approx 354 metres
Ignoring air resistance (the guy's only interested in a rough value) then the distance in metres =0.5 x 9.8 x time (in seconds) x time(in seconds).
So if it takes 10 seconds to hear the thud, then the distance is
0.5 x 9.8 x10 x 10=490 metres.
As factor says, sound takes a second to travel 300metres so we should subtract about 1.5 seconds off this time giving approx 354 metres
The rate of acceleration for a heavily loaded rucksack could easily be reduced by half, at the speed obtained in 5 seconds, and that's after falling only one tenth the total estimated distance.
http://www.arachnoid....s/freefall_graph1.gif
Assuming an average speed of 165 mph, the sound at impact from 3/4 mile below would arrive about 20 seconds from time of release.
http://www.calctool.o...ng/aerospace/terminal
In the situation described I would tend to presume no in depth calculations were performed. The distance estimate given was most likely based on previous times required for the rucksack with similar contents to fall a similar distance with perhaps a bit of interpolation factored in along with accommodations for a reasonable margin of error. Then again . . . it was a movie. ;o)
http://www.arachnoid....s/freefall_graph1.gif
Assuming an average speed of 165 mph, the sound at impact from 3/4 mile below would arrive about 20 seconds from time of release.
http://www.calctool.o...ng/aerospace/terminal
In the situation described I would tend to presume no in depth calculations were performed. The distance estimate given was most likely based on previous times required for the rucksack with similar contents to fall a similar distance with perhaps a bit of interpolation factored in along with accommodations for a reasonable margin of error. Then again . . . it was a movie. ;o)
Since it is practically impossible to know the terminal velocity of the rucksack the time of the fall would be inaccurate.
A better means would be to watch the rucksack explode when it hit the ground then measure the time for the sound to be heard. Sound travels at about 340 metres per second (about five seconds per mile) at sea level and room temperature.
A better means would be to watch the rucksack explode when it hit the ground then measure the time for the sound to be heard. Sound travels at about 340 metres per second (about five seconds per mile) at sea level and room temperature.