Terminal velocity 

Thank you. Love this comment. Cheers, Dr. A

The fact that he has a PHD before his name gives it away, but I understood your query.

@@yoprofmatt thanks for teaching the subject :) i couldnt find any video that catched my attention and manadged to keep it for longer than 2 minutes in portuguese before realising "oh wait,im bilingual and theres probably more videos in english than in portuguese" so i searched it up and am now seing your video and being able to learn the subject and i hope this will be enough to make up for me distracting myself and falling behind on class :)

@@yoprofmatt WHY AND HOW what are OBJECTS MAY FALL AT THE SAME RATE IN FULL CONFORMITY WITH WHAT IS E=MC2: CLEARLY, this is a gigantic breakthrough in physics AND mathematics. WHAT IS E=MC2 IS dimensionally consistent. TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE, AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE) !!! GREAT. WHAT IS GRAVITY is, ON BALANCE, an INTERACTION that cannot be shielded or blocked !!!! MAGNIFICENT ! Consider what is THE EARTH/ground ON BALANCE !!! Consider what is THE EYE ON BALANCE !!! Now, consider what is the TRANSLUCENT AND BLUE sky. CLEAR water comes from what is THE EYE. (THE EARTH is ALSO BLUE.) Fantastic. The stars AND PLANETS are POINTS in the night sky. BALANCE AND completeness go hand in hand. It ALL CLEARLY makes perfect sense ON BALANCE. INDEED, LOOK at what is the orange (AND setting) Sun !!! GREAT !!! Consistent WITH WHAT IS E=MC2 (AND WHAT IS TIME), “mass"/ENERGY involves BALANCED inertia/INERTIAL RESISTANCE consistent WITH/as what is BALANCED electromagnetic/gravitational force/ENERGY; AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE) !!! (BALANCE AND completeness go hand in hand !!!) Great. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE (ON BALANCE), AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE; AS ELECTROMAGNETISM/energy is CLEARLY (AND NECESSARILY) proven to be gravity (ON/IN BALANCE) !!! This IS consistent with WHAT IS E=MC2 AND F=ma !!! (ACCORDINGLY, ON BALANCE, the rotation of WHAT IS THE MOON matches the revolution.) Indeed, inertia/INERTIAL RESISTANCE IS proportional to (or BALANCED with/as) GRAVITATIONAL force/ENERGY; AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE; AS ELECTROMAGNETISM/energy is CLEARLY (AND NECESSARILY) proven to be gravity (ON/IN BALANCE) !!! This IS consistent with WHAT IS E=MC2 AND F=ma !!! GREAT. ❤️❤️❤️ I have mathematically unified physics. FACT. THINK. By Frank Martin DiMeglio WHAT RELATES GRAVITY TO WAVE/PARTICLE DUALITY?: WHAT IS E=MC2 does. TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE, AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE) !!! ACCORDINGLY, ON BALANCE, the rotation of WHAT IS THE MOON matches the revolution !!! The stars AND PLANETS are POINTS in the night sky. WHAT IS GRAVITY is, ON BALANCE, an INTERACTION that cannot be shielded or blocked. CLEAR WATER COMES FROM WHAT IS THE EYE. Consider what is THE EYE ON BALANCE, AND consider what is the TRANSLUCENT AND BLUE sky. Consider what is THE EARTH/ground. NOW, consider WHAT IS the ORANGE (AND setting) Sun ON BALANCE !!!! Great. THE EARTH is ALSO BLUE. Consider what is the BLUE flame. WHAT IS E=MC2 IS dimensionally consistent. Magnificent❤️. That what are OBJECTS may fall at the SAME RATE is FULLY consistent WITH WHAT IS E=MC2 !!! GREAT !!! TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE, AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE !!! Consistent WITH WHAT IS E=MC2 (AND WHAT IS TIME), “mass"/ENERGY involves BALANCED inertia/INERTIAL RESISTANCE consistent WITH/as what is BALANCED electromagnetic/gravitational force/ENERGY; AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE). Accordingly, ON BALANCE, what are OBJECTS may fall at the SAME RATE in FULL conformity WITH WHAT IS E=MC2 !!! Great. I have revolutionized physics. GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE, AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE, AS ELECTROMAGNETISM/energy is CLEARLY (AND NECESSARILY) proven to be gravity (ON/IN BALANCE). This is consistent with WHAT IS E=MC2 AND F=ma. Accordingly, ON BALANCE, the rotation of WHAT IS THE MOON matches the revolution. Accordingly, ON BALANCE, what are OBJECTS may fall at the SAME RATE in FULL conformity WITH WHAT IS E=MC2 !!! Great. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE (ON BALANCE), AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE; AS ELECTROMAGNETISM/energy is CLEARLY (AND NECESSARILY) proven to be gravity (ON/IN BALANCE) !!! This IS consistent with WHAT IS E=MC2 AND F=ma !!! I have surpassed Newton and Einstein. By Frank Martin DiMeglio

if you think he is writing backwards then he isn't. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-eVOPDQ5KYso.html ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-CWHMtSNKxYA.html&feature=emb_title the teacher is writing normally like everyone. He just somehow flips the video, in this case very likely with video recording. If you do not believe, look at the professor. most of the teachers who write on glass seem to be left handed. But statistically mostly are right handed. So this means that the video has been flipped on the vertical axis, so in this way the words that look opposite look normal. While the teacher image will also be flipped, since he is inside that video or image. This is why a right handed person looks left handed and vice versa.

that's why he's the professor

@@AkashdeepSingh-qq5fw I learned something today. Thank you.

So his class just has to read backwards… unlucky

Lol, right???

Connor, Talking to you, Connor! Too funny. I should add more names in randomly. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A

make sure to make the elephants as aerodynamic as possible, will increase terminal velocity

Don’t ya just hate people who say lol ALL the time? Btw, Felix Gonzales, nice profile pic 🙈

my 2nd Remote (Online) Physics exam brought me here! But I was thinking the same thing.....(altho honestly, I'm nust sick of trying to learn Physics online... and we're not even half way thru the semester😭)..... where do you teach, Professor?

Same brother

Thanks, I think so too. Cheers, Dr. A

Axel DIMBLAD, Great comment, thanks. You might also like my new website: www.universityphysics.education Cheers, Dr. A

FNAJ Development, Thanks for the comment, and yes it could use some updating. You might also like my new website: www.universityphysics.education Cheers, Dr. A

asked the same question by mail.. please share ans once you get

I believe it has to do with the Buoyancy of the objects. In a vacuum, they absolutely would hit the ground at the same time, any 2 objects will regardless of mass or shape. They did this experiment in a giant vacuum chamber with a bowling ball and a feather. But in the case of a balloon or basket ball, the air inside it makes it buoyant, which is a force preventing it from falling as fast, as the denser bowling ball. The basic principles of fluid dynamics for water and air are the same, except the density of the environments are different. Imagine then, dropping the bowling ball and the basket ball into a pool, the bowling ball would sink, but the basket ball would float, despite being the same shape and despite the forces of gravity pulling them downward at the same constant force. This is because the basket ball is more buoyant than the bowling ball. The same thing happens in air, but to a lessened degree since air is much less dense than water. This is also the same reason why helium balloons float, despite the fact that both the balloon and the helium inside it have weight and therefore should be affected by gravity. But the buoyancy of the helium compared to air, creates a force greater than the downward force of gravity. Thereby pulling it up, instead of down. Floating, instead of sinking.

@@superbart3000 Firstly I'd like to thank you for taking your time and answering my question with such detail. I didn't take into account the buoyancy and I also believe it has something to do with that, but something still doesn't add up. You mentioned they'd fall and hit the ground at the same time in a vacuum but they would also hit the ground at the same time in absence of a vacuum (there's a video from Veritasium which shows this, though he is using a medicine ball instead of a bowling ball. Regardless, the priciple is the same). The reason for this is inertia as mention in the video. Is there something I'm not taking into consideration?

@@themeditatingdog6402 hmm, if they have the same drag and buoyancy isnt a factor. Then yes they should hit the ground at same time. On earth all objects fall at a constant acceleration of 9.8 m/s2 regardless of mass.

if it were just gravity they would fall together

bro acceleration is constant = 0, therefore velocity will not change it will be constant.

That's not what he's doing you moron. Use deductive reasoning.

Krisitak, You are very welcome. Cheers, Dr. A

In theory - yes, but one thing to consider is that the balloon has air inside it - so in a vacuum, that balloon would surely burst. Take a feather and a bowling ball in a vacuum - yes they will fall at the same rate. In fact, there is a video right here on RU-vid of a feather and a bowling ball being dropped in a giant vacuum chamber and they indeed accelerate at the same rate.

Then it slows down to terminal velocity assuming it has enough time to slow down that much before hitting the ground.

Soham Sharma, Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A

@@yoprofmatt thank you sir for such a fast and efficient reply but I still unfortunately didn't get my doubt clear. Please help

@@sohamsharma742 Go to his website and see if that clears it up for you.

Per Enflo, The ones in my audience were all volunteer, so they sort of knew I would call on them. But appreciate your comment. You might also like my new website: www.universityphysics.education Cheers, Dr. A

You are welcome. Cheers, Dr. A

yes as there is no opposite force to stop ya Newton's First Law of Motion kid

@@CPrasannaDarvekar i like how you wrote "kid" in the end and yes im kid

-CHARRED -, You will keep accelerating if you still have some sort of propulsion, like a rocket engine. And no you can never reach the speed of light because that would require infinite energy. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A

Matt Anderson So, even in vacuum, while I’m infinitely falling down by gravitation force (9,8) I will have some kind of terminal velocity, or I didn’t understand you properly?

@@imstuckwiththisusernamefor2190 You can't fall infinitely. You only fall if you are pulled by gravity. Gravity only pulls towards a large mass. To fall infinitely would mean the space between you and this large mass is infinite. One mass that has an gravitational pull strong enough to pull something from an infinite distance would turn the universe into a singularity, in which time and space wouldn't exist. You can't accelerate without time or space.

If there is no wind, you initially fall at an acceleration a = -g = -9.8 m/s^2. (The negative sign just means "down".) Once you start moving faster, you have to push air out of the way. This drag from the air lowers your acceleration until finally your acceleration is zero, hence terminal velocity. We only looked at the start (a=-g) and the end points (a=0), but the actual curve of your velocity is rather interesting. Hope this helps. Cheers, Dr. A

I don't know, what is sigma Fy? Cheers, Dr. A

That is the sum (sigma) of all the forces (F) in the vertical (y) direction

Yes. At terminal velocity (no more acceleration), the sum of all the forces in the y direction = 0. Cheers, Dr. A

Thank you so much sir!

You're very welcome. Keep up with the physics! Cheers, Dr. A

Great to hear. Cheers, Dr. A

Positive and negative doesn't matter really. You just need to be consistent with it. For a two-dimensional problem, where there is only up and down. Just make sure if you use a positive number for the "up" direction, any other vector which points upward is also positive - and anything downward is negative. Or vice versa. Acceleration due to gravity is downward. Drag is upward. Initial velocity is zero, but will obviously be downward immediately after the initial state.

+indra fitriyanto Excellent question. In our model, we are simplifying it to the linear case, which works well for small speeds. For higher speeds, the quadratic is more appropriate. You'll see both versions in different textbooks. Probably for a skydiver we should be using the bv^2 option. Thanks much, keep up with the physics. Cheers, Dr. A

John Sumner, Great point. Somehow air resistance didn't kick in enough for him to tell a difference. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A

a = 0 refers to the net acceleration, it means that g does not accelerate the object anymore because the deceleration due to drag is equal and opposite.

You're correct. Initially after you jump, vertical velocity increases for a while until the air resistance = weight. At that point, a = 0 and the velocity is "terminal" (does not change). Cheers, Dr. A

Galileo assumed it in vacuum, if u drop any object for instance a bowling ball and a feather in vacuum they will fall at exactly the same time. Now in practical world if u drop both u know, I know bowling ball will fall way faster and collide with the ground before the feather does, due to air resistance. By the way if you want to prove Galileo correct (LOL) just take a paper (exactly size of the note book) and a notebook . Place the paper just above the notebook then let it free fall.....AND abrakadabra they fall nearly in the same time. Reason - the air particles are displaced by the notebook for the the paper to freefall. (When I first tried it I was amazed) Just an Enthusiast !

You are right. But you have to taking into account buoyancy. Despite have the same shape and size. The basket ball is more buoyant, meaning because of its density compared to the air surrounding it, there is more of an upward buoyant force acting against it, canceling out some of the acceleration due to gravity. Imagine dropping both of these objects into a pool of water. The buoyancy force in that fluid environment would be so strong that the basket ball would not sink, yet the bowling ball would. Because the ball is filled with air which is less dense than water, and the bowling ball has a very dense center. The same thing happens in air, but to a lessened degree. The same thing also happens with helium balloons that would happen if you placed an air filled balloon at the bottom of a pool of water. The buoyant forces would pull both balloons upward, because of the density difference between the balloons and their environments.

Depends on the height, there is a certain high where weight acts, if you drop those on a height of 2 meters they will fall at the same time

muhammed akram, Remember that + and - signs just indicate direction and that direction depends on your coordinate system. If you draw your coordinate system with +y going up, than a = -g. If you draw your coordinate system with +y going down, then a = +g. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A

I believe the negative just shows the direction in which gravity acts.

You can set any direction to be negative as long as everything is consistent

Not really. Free fall is motion free from all forces - except gravity. Perhaps in the first few seconds of a skydiving jump, the drag is negligible and you are very close to free fall, but technically not. Near the Earth's surface, the gravitational acceleration is 9.8m/s², so when you first jump out of a plane, you accelerate at this rate, but almost immediately, drag is acting on you - which is a force in the opposite direction. This counters the force of gravity, so this net acceleration decreases. 9.7 m/s², then 9.6 m/s² and so on. The faster you go through the air, the higher the drag force becomes, so the more and more it counters gravity, so just before reaching terminal velocity, your net acceleration is 0.3 m/s², then 0.2 m/s², 0.1 m/s² and then finally - 0.0 m/s². You are now going so fast that the drag force equals the force of gravity, so acceleration stops. You will no longer get any faster. This is terminal velocity. You don't go from free fall (which is 9.8 m/s²) directly to terminal velocity. Your net acceleration gradually decreases as your speed increases (which is increasing the drag force - until it equals gravity).

The important detail is that there is air resistance. In the absence of air resistance, Fd would be zero and both objects would continue to accelerate at exactly 9.8m/s^2 and hence hit the ground at the same time.

Inside a vacuum both balls will hit the ground at the same time.

Nope. Cheers, Dr. A

why would it be?

@@willoughbykrenzteinburg deep.

Oh wait, it's not? No wonder the graphics seemed really good today...

Unless you have a death wish.

Good day sir, I so much love this video. Can you please teach me how make a video like this including devices needed for a production like this. Thanks

Good question. Remember that mg is a force. What this means is gravity is pulling on the object with a force equal to mg. If no other forces are present, then the mass m will accelerate at g because ma=mg. This is what happens when the skydiver first jumps out. However, with other forces present, the acceleration can be very different, since ma = mg - Fdrag. Thus if mg = Fdrag, then the acceleration is zero. This is when the skydiver reaches terminal velocity. Cheers, Dr. A

g is what it always was. a = 0 because F = 0. It's important to understand that this does not mean that there are no forces acting - it just means no *net* forces are acting. In other words, all forces cancel each other. Particular for something at terminal velocity, a = 0 because Fnet = 0. Fnet = Fgravity + F drag. If the two are equal, then the net force is zero - because they act in opposite directions. If the force of gravity is 50 Newtons (downward), then the force of drag is 50 Newtons (upward). The net force is therefore zero, so a = 0.

Remember that if the net force = 0 then there is no acceleration. But of course there can still be velocity. Remember that acceleration = change in velocity. When mg = Fdrag, the acceleration is zero which means the velocity must be constant. You're still falling down, but at a constant speed (and direction). This is what we call terminal velocity. It doesn't change. Hope this helps. Cheers, Dr. A

I have also the same question. Mr. Anderson, Would you please help us.??

what>?

Nope, it's not the same

It doesn't really matter. It's just the designation given. If all values that are presented as positives were negatives and all values that were negatives were positives, you'd arrive at the same result, so it's arbitrary. But generally, down is negative and up is positive. g is an acceleration in the downward direction, ergo it's generally represented as a negative.

Uranium Studio, Great to hear. Who's John Petrucci? Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A

Nope. Secrets here: www.learning.glass Cheers, Dr. A

Good catch, but not true. Check us out here: www.learning.glass Cheers, Dr. A

Wow that's really cool. Always fascinated by things like that.

Stokes law only valid for spherical objects, in fluids..............It is similar to air drag but not air drag

Depends on coordinate system we choose

Patrick, I completely agree. There is much more to this story that we need to develop. We're working on it. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A