Differential Equation - 1st Order Solutions (8 of 8) How to Calculate Parachutist's Terminal Speed

[@rodericksibelius8472]

This is one great thing about your KZbin Channel Professor Michel van Biezen, Your teaching is so detailed, explained in simple terms without too much jargon that even elementary and high school graders can follow the way you TEACH. Thank God We have a Teacher LIKE You. God Bless. Keep Up the GREAT WORK EDUCATING the World of Physics and ENGINEERING.

[@MichelvanBiezen]

Wow, thanks! We appreciate you comment.

[@Shark323-h3e]

Excellent, impressive and an outstanding approach, that was really a remarkable way to explain differential equations in an interesting way. Thank you, Professor,, Michel.

[@alicandagasan6921]

Question 1: With its equipment, a parachutist with a total mass of 80 kg easily opens right at the first angle of 15 m / s at a vertical angle from a stationary helicopter 1000 m high from the ground at time t = 0. The parachute is immediately opened at the time of the jump, and by moving it, a friction proportional to the speed of the paratrooper, with the proportionality coefficient k = 30 kg / s. a) Mathematical equation has been obtained where you can find the change of the speed of the paratrooper over time. b) position-time equation is obtained from the velocity-time equation. c) Similarly, the acceleration-time equation was obtained from the velocity time expression you found. thank you :D

[@Zeusbeer]

For the integral (1/(x^2-a^2)) there is a very neat identity that this is the same as 1/a arctanh (x/a) (hyperbolic arctan). Which is almost the same as for the identity 1/(x^2+a^2) which is (1/a)arctan(x/a) and thus quite easy to remember. This saves a lot of thinking on this exercise.

[@MichelvanBiezen]

Great input! Thank you.

[@lordyabo9910]

Another excellent playslist! Thank you again Sir, and God bless!

[@MichelvanBiezen]

Thank you. Glad you are enjoying the videos. 🙂

[@rashmiranjannayak3251]

Perfect way to find the solution; and much clarity in process of solution. Thanks for sharing.

[@curtpiazza1688]

Wow.. Great problem! I see the importance of studying your algebra and pre calc!❤ 😊

[@MichelvanBiezen]

Algebra and pre-calculus are foundational to many sciences.

[@ginginthing]

Too late. The jumper was trying to do these calcs and completely forgot to pull the rip chord. Guess he reached his final "terminal" velocity.

[@abdallahhussaini5847]

Wow!! Simply amazing !! I was stuck on the differential part for 4 hrs and you Sir have saved me from killing myself!

[@MichelvanBiezen]

HaHa, we know the feeling. We were students once (a long time ago).

[@CRohitSorte]

@@MichelvanBiezen Sir during integration why we dont put limits on v from 0 to V on t from 0 to T. If you do so while substituting for lower limit for v you will get log(-1) Which is invalid You try it sir

[@CRohitSorte]

Oh sir my fault It is the modulus of log after integration which I forgot to use :)

[@QueenL]

Yeah me too 😂😂

[@yonatanshenhav1208]

a very good video, and I can assure you I never say that about any other videos. thank you for making it

[@jaiakash3819]

Thank you ☺️ very much, it helped me a lot in my class today

[@MichelvanBiezen]

Glad you found our videos and you find them helpful. 🙂

[@Landrytommy]

To determine terminal velocity, can you go straight to solving mg-bv^2=0 ... Since terminal velocity is when acceleration is 0. In that case, i wouldn't benefit of knowing velocity at each point of time..

[@MichelvanBiezen]

Try it and see if you get the same answer.

[@Landrytommy]

I did, it works. I was just saying so to keep people talking...

[@Landrytommy]

Sqrt(mg/b)= 3.96

[@MichelvanBiezen]

Thank you

[@mariussundesivertsen5870]

Nice example and nicely solved, just as usual! You could have found the terminal velocity by setting the sum of the forces equal to 0 and solve for "v", if that was your prime objective as the title says :)

[@MichelvanBiezen]

Marius Sivertsenen Actually the prime objective is to illustrate a particular technique. Many problems can be solved in various ways

[@abdallahhussaini5847]

@Michel van Biezen, Sir, why dont you take the absolute of natural log as per the integral formula?? If you do so ln|v-k| would be -ln(v-k) since v will always be less than k which is the terminal velocity. Your explanation would be highly appreciated.

[@MichelvanBiezen]

In mathematics the absolute value symbol is used for the ln(expression) after integrating as a general rule in case the expression is negative, since the ln of a negative number is undefined. The absolute value is not a requirement or neither does it need to be there every time. Only place it there if the physical situation requires it.

[@ing.franciscocabrera.msc.5957]

Hallelujah excellent demonstration .. !!!

[@dragonore2009]

Could you explain why you took the limit as t approaches infinity better? Was the reason mainly for a sanity check to confirm everything worked out?

[@MichelvanBiezen]

We let t approach infinity to show that terminal velocity will equal k.

[@dragonore2009]

Oh, okay, thanks Mr Biezen. Also I wanted to say I appreciate your videos, it really helped me out during my physics class. I'm sure when I take physics 2 I will be be right back to your channel again.

[@secretpassion3801]

Superbbbbbb....i loved it!Thank you Sir

[@scar0134]

Is b =. 5*density of air*area of the parachute*drag coefficient of parachute? If that's the case, the units are kg/m and that is not the units he has for b, How do you find b?

[@muhammadhabibannajar1115]

scar0134 it's a constant depending how dense the air

[@EDUARDO12348]

Thank you for great videos. On this one am struggling to understand the units of k. From what I see online, b = Ns/m^2

[@MichelvanBiezen]

On the left side of the equation, the units of each term is "N", which means for the units of "b" to cancel out the units of "v^2" the units of b must be N sec^2/m^2

[@EDUARDO12348]

Wow thank you for reply! may I ask one more question? Do we need to convert Newtons to kg for k?

[@MichelvanBiezen]

Since we use k as a temporary substitution to make it easier to work the problem, we don't have to worry about the units, and since k does not represent anything physical, just a convenient substitution.

[@LuisLopez-nl6qv]

What exactly is it that he did at 4:00 ?? I still don't understand it

[@sideriaknight6338]

Question, how could we use the final equation we got of velocity in function of time to know how long was a man in the air if the height he jumped from were 3000 m?

[@MichelvanBiezen]

Distance is represented by the area under the velocity curve. Thus you would first have to integrate it to find the distance traveled as a function of time.

[@chifungok]

Why did you reverse the (k^2-v^2) at 2:20?

[@MichelvanBiezen]

+Chi Fung To put into a standard form (variable first)

[@chifungok]

Thank you :)

[@forasago]

I guess I'll watch this in a few decades when I can no longer hear the high pitched noise.

[@GrassBIade]

hah, I didn't even notice.. not sure if I'm glad or not

[@inaki7233]

How old are you? I'm seventeen and wondering when will I stop hearing that.

[@GrassBIade]

late 20’s. Pro tip- don’t join a punk band

[@mattkazmer451]

I like your style, very nice!

[@MichelvanBiezen]

Appreciate the feedback. Glad you like it. 🙂

[@ptyptypty3]

WOW.... perhaps the Jumper should have used a Hang Glider !!!.. lol... Congrats Michel!!. You just strapped on a 100 pound backpack and jogged through 2 miles of Knee Deep MUD and made it to the finish there!!... WOW... Great Problem and Presentation!!... thanks!

[@MichelvanBiezen]

Differential equations can be fun.........:)

[@于洋-q5m]

really clear process, i like this way, nice!!!!🥰

[@MichelvanBiezen]

Glad you liked it!

[@LatinoPrototype]

Wouldnt it also be possible and perhaps even easier to solve the DE using an integrating factor?

[@calseynez]

I have a college question, what course levels do you teach and what subjects do you cover for each chapters. and thank you for the video now I understood the concept and steps of falling bodies and air resistance. do you have another videos for bungee jumping and some other videos for statics and dynamics.

[@ajaytak7488]

can you plz tell how are we gonna know whether the parachutist will attain the terminal velocity or not, because the height at which the parachutist opens his/her parachute may be too small. so can u tell are can we find out that ??

[@peterkovacs1933]

Hey could someone explain what the constant b is? What is its name and how do you find it?

[@MichelvanBiezen]

Whenever there is an object moving through a fluid you will have a drag coefficient that depends on the shape of the object. It appears in the equation as a first order term in the differential equation and the drag force associated with it usually (depending on the flow) depends linearly on the velocity.

[@kaholeung9443]

Thank your sir , very clear steps

[@MichelvanBiezen]

You are most welcome

[@CaptainCalculus]

What would the c=0.433 represent?

[@LeonardoLoboEngineer]

Michel, when t goes to infinity, in the final step the exponential part of equation goes to 0, so whe stay with Vf=3.96 x 1/1 whitch is = to final velocity = to 3.96 m/s ... did i make something wrong? sorry by my english! thanks for this ODE series!

[@MichelvanBiezen]

Leonardo Lobo That is correct. The final velocity is 3.96m/sec. (He opens the parachute when moving at 10m/sec and slows down to 3.96 m/sec.

[@lillyzegarra8025]

whatever happened to the +C in (1+C)? How did you make it disappear? It make no sense to me, there's a C on the right side and a C on the left side, how did the C on the left side disappear?

[@MichelvanBiezen]

If you have a constant on the left side and a constant on the right side you can combine them and call them a constant on one side. Remember that "c" represents any constant.

[@lillyzegarra8025]

@@MichelvanBiezen I wish you would've clearly stated this during the tutorial and the reason I say this is because those of us who are watching already know that a C in the right is to be combined with the C in the left but we still need to hear it from you, don't ask me why? Sorry about the old joke, I didn't mean it and thanks a lot for taking the time to explain the things that go on when you block the board.

[@MichelvanBiezen]

Usually I do mention that. May have skipped it there.

[@cjoutdoors1991]

Wait, is g positive? It seems like letting k^2 = (m/b)g (keeping in mind that m/b is positive) means that this must be true.

[@MichelvanBiezen]

It depends on how it is used or how it is defined. In this case we are looking for the magnitude of the velocity as the object moves in a downward direction.

[@lillyzegarra8025]

How do you get 1.525 on the left side, it is a mystery, you were blocking the board and I couldn't figure out how you rid yourself of the C.

[@MichelvanBiezen]

2.525 - 1 = 1.525

[@lillyzegarra8025]

Get out of the god damn way, stop blocking the white board. You're not the only one using a white board and everyone else is way more considerate than you are. Don't tell me you're to old to learn new tricks. Great class by the way, i enjoyed it, you rock dude.

[@MichelvanBiezen]

No reason to take God's name in vain. But yes, this was a problem we were struggling with , which we eventually solved by working from the right side of the board to the left side and by paying special attention where I stand. Part of the learning process.

[@วิชยุตม์-นาคะศูนย์]

When you got the integration both side,why the left side wasn’t plus the constant sir?

[@MichelvanBiezen]

You can simply add the 2 constants on both sides and make it one constant on one side.

[@วิชยุตม์-นาคะศูนย์]

Michel van Biezen Aww it’s C1. Thank you professor.

[@bonnehh_4301]

how did you assume the b value as 50?

[@MichelvanBiezen]

You can look those up on a table, but I believe I just made that one up.

[@dpmike32819]

The math trick at the 3:53 mark

[@ndi_wpc]

i cannot solve the next form h vs t, and h vs v..i just want to estimate the velocity when the body touching the ground

[@MichelvanBiezen]

The value of the constants are given in the problem. You just need to substite them into the final equation and you will get your answer.

[@Saurav.nittian]

A man of 80 kg.falls freelly from height1000m. upto 500m.he falls freely then retarding and reaches at earth with velocity 10m/ s .calculate work done by resistive force​ Please solve this

[@MichelvanBiezen]

If there was no retarding force the velocity would be Vf = sqrt ( 2 g h) = sqrt (2 x 9.8 x 1000) = 140 m/sec Instead the final velocity = 10 m/sec The work done by the retarding force equals the difference in KE. Work done = (1/2) (80) (140)^2 - (1/2) (80) (10)^2 = 784,000 J - 400 J = 783,600 J

[@mr.papalavaskia9795]

Your scaring me dude

[@ardisyahda1209]

Thanks for sharing

[@SparkTechUg]

Thanks very much

[@MichelvanBiezen]

Glad it was helpful. 🙂

[@saminzawad4677]

Please recheck the unit for viscosity. I think it should be N.s/m^2

[@MichelvanBiezen]

Each term in the equation must have the same units (N). b represents the drag coefficient and therefore has the appropriate units.

[@วิชยุตม์-นาคะศูนย์]

How to find the terminal position sir?

[@MichelvanBiezen]

Since the parachute keeps moving there is no terminal "position".

[@วิชยุตม์-นาคะศูนย์]

Michel van Biezen Oh yeah I mean position in function of time. I want to find x(t) from F(x) like this where x F t is position,force and time Respectively. May you suggest please?

[@emmalucas5379]

Oh god that insanely high pitched hum in the background is like a human dog whistle

[@aminulislamstudent5771]

thanks a lot

[@juanpablorequenamarquez9839]

I would have líked to understand, but you sir write so weird