When Is A Bungee Jumper's Acceleration Max?

[@veritasium]

well said. Just to be sure I counted the frames till the rope was taut, and then counted the frames from there to the bottom. There are more frames accelerating than decelerating which also supports the conclusion that the magnitude of acceleration is greater than g at the bottom.

[@karansinghmadhwal1074]

i am correct

[@karansinghmadhwal1074]

a point d this is depend upon on the mass of object sir

[@sudarshan1793]

acceleration does not change in both up and down motion

[@idsrevenge]

Great discussion piece for teachers who are working on the motion of springs. Extremely challenging to interpret this for physics students who've only worked with kinematics, maybe a sort of challenge question, or something for kids in AP. Thanks for putting this out there, explanations for each option are great.

[@veritasium]

Here's the thing, kx is not independent of mg. In order to slow down, kx must be greater than mg, then you need to keep extending the spring until you stop. By this point kx will be greater than 2mg. Proof: set gravitational potential energy = spring energy at full extension: mg(l+x) = 0.5*kx^2 [x is the extension, and l is the length of unstretched chord] then 2mg=kx - 2mgl/x; now if l=0, 2mg = kx but if l>0, kx must be bigger than 2mg

[@luckypegasusvol7700]

Why 2mg?

[@래모루래모로]

@@luckypegasusvol7700 2miligram

[@sars5813]

Random question but as you’d mentioned in one of the videos you aren’t accelerating at all are you?

[@exon901]

Very creative thinking used in this video. Instructive, thought provoking, a real mental challenge for those of us who did Physics many decades ago. Liked the use of multiple choice answers, as well as the graphics and succinct explanations. especially velocity vs acceleration. Look forward to more from you Derek. Well done !

[@veritasium]

This is a good point. I made the videos public so viewers didn't have to come back to this vid to get the link to different options.

[@alexcrack5442]

Is letter C, because it’s negative acceleration

[@prateekpanwar646]

@@alexcrack5442 D precisely

[@Zeyev]

Really well done. I watched all of them to more fully understand what you were teaching. It has been only 45 or so years since I had a physics class so it's a great reminder of how much I have forgotten. Thanks.

[@amitintelegent5913]

acceleration means rate of change of velocity there for point 'a)' must be highest for rate of change of velocity with respect to that short time interval

[@ashutoshsharma8334]

a) the acceleration is g + the tension in the hanging rope. (between a and b, tension is reducing) b) Tension has just started to oppose gravitational acceleration. c) highest speed, means that equal accelerations in opposite directions. (tension - gravitational acc) (between c and d, tension goes on increasing until point d) d) Tension is now the greatest and is opposing the gravitational acceleration as much as it can. e) Tension has again started to favour g. Therefore my answer goes to "a", because I believe That even though tension in the case of "d" is opposing acceleration as much as it can but its magnitude can only become equal to that in the case of "a", and since in option "a" tension and g are added while in option "d" they are subtracted, acceleration produced in the case of "a" is more.

[@Error-yh3xr]

I'm sorry but why would there be tension in a) when the rope is not taut yet?

[@waroftheworlds2008]

@@Error-yh3xr the tension would be zero....ish. half of the rope has tension.

[@luckypegasusvol7700]

@@waroftheworlds2008 what is the actual answer?

[@waroftheworlds2008]

@@luckypegasusvol7700 i totally recommend watching the video.

[@imaginarycastle]

I'd recommend making the different answer videos hidden from public view because of two reasons: a) Some people find and watch one of the answers first, not knowing what it's about, and thus missing out on the whole thing. b) Since the different options appear as related vidoes, I think it influences the choices made by viewers. If one of the options list higher than the others or have more views, people might choice that option for that reason rather than trying to answer the question at hand.

[@CrisHaasbro]

The accelleration of gravity is a contant 9.81 m/s^2 however his actual acceleration being influenced by friction and the spring effect of the bungee is not constant. His greatest accelleration would be at the top of the jump before air drag and the spring effect of the bungee counter it and once again after bouncing up to fall back down again.

[@azuma3221]

D because...Your point of greatest acceleration is actually at the bottom, where your velocity is momentarily zero, but changing very fast. That's also the point where the force on you is greatest.

[@LeavingCertMaths]

We do not have to assume an elastic force law of the form F=kx. We just need the reasonable assumption that the elastic force increases as the stretching increases. Before the rope is stretched, the magnitude of the resultant force on the man is mg. After the rope is stretched, the resultant force is F - mg where F is the elastic force. When the man stops, the stretching is maximum and hence F and F - mg are maximum. Therefore by Newton's second law the acceleration is maximum when the man stops

[@ArmednSafe]

Not enough information. If the bungee has a very low elastic coefficient it will be during the free fall. If the elastic coefficient is very high, it will be at the bottom.

[@checklinderchap7570]

Technically if an object is free-falling, it must have only ONE force acting upon it: Gravity (for example no friction from air or other)

[@ArmednSafe]

Checklinder Chap You're right. I was using the term "free fall" loosely.

[@danielrhouck]

Given only the words in this vide, I believe you're correct (unless you were to look up the spring constant of typical bungee cord). However, we have a video to go with those words, which does give us enough information: he fell a shorter distance between points c and d than between points a and b, and he wasn't even done accelerating downwards at b. Thus, he must have had a greater acceleration on average between c and d then he did before the rope became taut.

[@ArmednSafe]

I don't see any calculations. Show me. How long is the bungee cord? What's the elastic constant you are using? What's the mass of the jumper? If you stick in values for those things, you can get an answer for peak acceleration. But, then you have to do sensitivity analysis on those numbers you stuck in to determine how much error your estimate can tolerate. Does the answer change if you're off on the jumper's weight by 20%? 5%? Veritasium is just some guy with a camera. He is fallible. This video is nonsense.

[@Dowlphin]

This is indeed very vague. There are several factors involved, but it is difficult to figure out how significant they are. After jumping off, his movement is mor an arch and not straight downwards where it would create the highest acceleration, before air resistance slightly reduces it. But he also begins with his body flat in the air flow, thus further increasing the friction. When the rope starts to slow him down, this could be the greatest (negative) acceleration, but does a negative number count as a maximum?

[@TheOnlyRizzy]

It's either A or D, depending on the properties of the rope. At A, his acceleration is 9.8m/s^2 downwards. At D, his acceleration is the difference between the upwards acceleration of the bungee cord and the downward acceleration of gravity. So if the rope can provide an acceleration greater than 19.6m/s^2 upwards, then the magnitude of the net acceleration is greater than 9.8m/s^2. However, if the rope can't provide that great of an acceleration, then the answer remains A.

[@DarkMagician903]

Acceleration is defined as the change in velocity with respect to time. Considering at D the velocity literally changed directions from towards the Earth to away from the Earth, that is the point of the largest change in velocity, therefore being the point of most acceleration and force on the bungee jumper.

[@aeonian_harshul]

But at that Instant he is at rest, hence by Newton's 2nd law the net force on him must be 0 and hence the acceleration

[@redShiftish]

This is a common misconception of Newton's laws. No law states that an object at rest has no forces acting on it. The first law states that an object a constant velocity has no forces acting on it and vice versa. The jumper is not at a constant speed at the bottom; his speed is changing rather quickly like Oliver said. Therefore, the forces on him are large. Newton's second law says that the total force is proportional to the acceleration and, while relevant to this situation, is not relevant to the specific question of whether there are forces acting on him.

[@speedstyle.]

Imagine you start your car, reverse out, then accelerate down the road into a brick wall. You change directions in the parking lot, but clearly the largest acceleration and force is at the wall. Going from -10ms⁻¹ to 10ms⁻¹ in a second is the same acceleration as 100ms⁻¹ to 120ms⁻¹, the change in sign doesn't matter. You can also see this by considering a different frame of reference: the earth is moving about 30 000 ms⁻¹ around the sun, why would 29990 to 30010 be a particularly large acceleration?

[@shikumei5267]

We watched this is Physics and my friend and I just died laughing 😂

[@Laughing_Cat_Meme]

Imao😂😂

[@kayg750]

The answer given assumes the rope accelerates the jumper at a rate greater than 9.8m/s². You could have a really stretchy cord that would very slowly change the jumper's velocity. In that case, the jumper's acceleration is at a maximum right after jumping.

[@LeavingCertMaths]

If F is a force of the form F = kx^a where a >=1, then your energy analysis will also give F > 2mg (and hence maximum acceleration) when the man stops momentarily.

[@phk999]

during the rebound i.e., the case 'E', the body does not go up much. because most of the negative acceleration developed in the cord(tension) is expended at 'D'. after the cancellation only a little more deceleration left for the rebound.

[@ShrubOfDeath]

D. Acceleration is change in velocity over time. That change in velocity (directional speed) is highest at the bottom. Its the seemingly most "violent" change in speed.

[@sid2567]

Hi

[@Maninawig]

I think he FEELS acceleration the most between A and B, but he might feel it's effects more at D.

[@guuu881]

The force of gravity is basically constant, but the other forces acting on the jumper vary over the course of the jump.

[@tonyli8368]

Um, shouldn't it depend on the amount of tension the rope has?

[@jonahansen]

Totally depends on the rope and whether it obeys Hooke's law.

[@rahuln5676]

Finding this comment a year later, so underrated comment lmao!!

[@meesamtamar3650]

" A "as some of the energy will be lost when the string steaches and then transfers the energy to the man just like a pendulem the higest acceleration of the bob is when you leave it and it starts the oscillation and its maximum speed is at point b where it is 90 digrees with the floor and and then the acceleration is greatest at point "C" although the acceleration being a vector quantity will have its direction (opposite) and almost the same as at point A but here it is not a bob and its not a perfect oscillation theres some energy wil be lost when the string pushed the man upward (please correct me if i am wrong 😁)

[@Geophrie39]

Nice! "A" is in fact, the correct answer

[@polatiger4765]

Depends on the rope. Just before B if the rope is very elastic. (E is higher than B.) D if it is not so elastic.

[@MRHumourMnr]

The acceleration is same or constant throughout the whole jump because when we free fall the acceleration of a boby or an object is equal to the gravity is equal to 9.8m/s2.

[@phk999]

'D' it is. At 'A' the person jumps, his velocity is the minimum but the acceleration due to gravity acts. At 'B', he still accelerates, because to reach the terminal velocity he has to jump from a height of 381m nearly(size of the empire state building). After 'B' until 'C', he reaches maximum velocity, but the tension in the rope builds up. And at 'D' i.e., the bottom of the jump, the tension in the rope becomes so high, it can counter-act the acceleration due to gravity and stops the body from moving down further. In addition it pulls him up. This deceleration is the maximum. So the answer is 'D'. the answer is the same for any kind of material used to make the rope.

[@markmongan8298]

D was my first choice in my head due to opposite force, but it was the way you said E. Tricked me again Veritasium!

[@archangel1996j]

the acceleration is constant only the velocity changes with time due to the acceleration , therefore he feels the same acceleration through out the motion with just different velocity

[@icesalesman2644]

Took me too long to realise that the acceleration is constant

[@adityakushwah4980]

How is it constant

[@gopeshingale4215]

its not constant brother

[@Hanuman120]

I imagine that this video used to link to other videos or are the annotations just not showing up on my device?

[@quasa0]

yep

[@quasa0]

got the links for you

[@evanb.6150]

I would say E: On the rebound. This is due to the fact that for the guy to be moving upwards, the rope has to be imparting an acceleration of a greater magnitude than gravity in roughly the opposite direction. Either E or B, I say.

[@poligon333]

Thanks for the video, well explained.

[@henrypereira1745]

D because of the energy stored into the bungee cord being released at it's maximum stretch.

[@lucaraujo1]

B, because the elastic force isn't pulling you back yet. The only force is gravity that is making you go down faster. But after B, the action of the elastic force begins.

[@jimharmon9917]

The answer is A - when acceleration equals G. At every moment after this, acceleration is reduced by air resistance and/or the elastic properties of the bungee cord. The velocity isn't at maximum until later, but the acceleration at that moment is less than G.

[@shgysk8zer0]

B. When he begins decelerating. It may be negative acceleration, but a - b is constant and the rest is going to be less. Another mobile user who cannot click on the links. Was hoping to see then in description or comments.

[@kazishadidraiyan]

You're right.

[@Pyotyrpyotyrpyotyr]

I found it helpful to think of it in terms of force. F equals M a and since his mass remains constant you can ask yourself when does experience the most force and it's at the bottom.

[@toastbrot2153]

No, at the bottom, the experiences no force at all, because the two forces (gravitational force and the force of the rope) add up to 0.

[@azlhiacneg]

2 years later... I watch this video. Could someone give a v-t graph of the bungee jumper? (A graph on a x-y plane with time being the independent variable and velocity being the dependent variable?) where going down is positive?

[@deday6525]

it's C. in order for the cord to pull you back, the tension must be greater than weight, acceleration due to tension must be larger than acceleration due to gravity.

[@arjanpolling6552]

My little try at this...Acceleration is always the same while your speed is changing. So when the rope is starting to suck up some energy (e.a. pulling you back) according to F=a*m > if F is less and m is the same...ergo... "a" must have lowered. So my answer is solution A.

[@arjanpolling6552]

:-( and :-) at the same time. This is why I love Physics!

[@jdgrahamo]

You should distinguish between acceleration and rate of acceleration. You also conflate acceleration and deceleration, which is confusing to say the least.

[@maurobruzzone5178]

does it has anything to do with the k (constante del resorte).... if the k was lower could it be option a? (sorry for my bad english)

[@danielrhouck]

I got this correct the first time. However, I have to wonder: if the jumper used a different type of rope or similar support, might the highest acceleration option and the second-highest acceleration option trade places? I *think* they would if you could get a loose enough spring.

[@LeavingCertMaths]

Perhaps you could give a justification for using energy conservation here. Gravity is not the only force that does work on the person falling.

[@Dixavd]

*Spoilers for anyone who hasn't picked yet - so pick already and stop reading my comment* I knew this but I understood what you said wrong so I clicked E I thought D stopped when he stopped at the bottom and E started immediately after wards but I knew that the acceleration was the changign of speed in whichever direction so the slowing to a stop and then increasing int he opisite must be the rate they experience the highest acceleration Nice video - I espially liked the comp. drawing graphs.

[@friggin149]

euhm, with acceleration you don't mean delta V right? because your differance in velocity is greatest when the cord tightens and you go in the opposite direction, completely negating the accumulation of velocity up till current velocity and adding a few extra m/s in the other direction. but that greatly depends on how hard the cord pulls. if you mean acceleration in falling it would be A where the air resistance is lowest and gravity is fully used.

[@funkyflames7430]

Given that the fastest point is above the lowest reach, must mean that point a must have been the time with the greatest acceleration. a = g b =

[@semicolon.d]

how do you know if the acceleration up at the bottom is greater than a G, i was hoping you would explain this but you didn't. if the cord can give the jumper more than a G of acceleration up then D is correct, but if not, the answer would be E(the point where the cord stopped pulling up), where the jumper is pulled down by the gravity plus the air friction, the accelleration would be slightly more than a g.

[@albirtarsha5370]

Mechanical energy is conserved in this scenario. Mechanical energy is kinetic energy (KE) + potential energy (PE). At the top (a) and the bottom (d), velocity is zero; so KE = 0 and the PE is the same at (a) and (d). Let h1 be the height from (a) to when the bungee is taught (b). Let h2 be the height from (b) to (d). Then PE at (a) = the gravitational PE = m*g*(h1+h2). From Hooke's Law, PE at (d) = the elastic PE = [k*(h2)^2]/2. Again ME is conserved so: m*g*(h1+h2) = [k*(h2)^2]/2. If you divide both sides by m you get that the acceleration due to gravity times some distance equals a greater average acceleration over a lesser distance. g*(h1+h2) = [k*(h2)^2]/(2*m) Then dividing by those distances: g < [k*h2]/(2*m). The instantaneous acceleration at (d) is ( (the bungee tension / m) - g ) = (k*h2/m) - g.

[@amaysharma1909]

No bro mechanical energy istn conserved in this case as when the string becomes taut there is some loss in energy due to the impulsive tension

[@JoeSmith-uo5yn]

I guess A .. 9.8 m/s My thinking is that acceleration is constant during freefall, and all the rest is a more gradual change in velocity.

[@ash4pm]

I agree - and because of that I think the question is erroneously phrased with no correct option for the answer which is that the acceleration is constant until the cable begins to slow him down. A better question would be: At which point does the jumper reach maximum speed? And the answer is probably just after the point at which the elastic cable resistance kicks in.

[@friggin149]

Just watched the answers and i think you're not completely right. As i said before it greatly depends on how hard the cord pulls, see it's possible that the cord pulls on you longer than you fell with a force only slightly stronger than gravity in that case the (delta V or acceleration) would be less than the acceleration due to "gravity at point A". ofcourse this is hard to simulate because you'd need a very strong and extremely elastic rope. ps. not talking to myself, just adding this.

[@vmadhavan435]

First ever veritasium styled video

[@somedayismarch21]

(I have two answers) Not 1 point ... but one bracket of time would have the highest acceleration From point A. all the way through point B. Should equal 9.8 meters per second. (assuming bungee stores less than enough energy to pull the man 2 G's upward from lowest point) All other points would be less because ... point C bungee resists. point d bungee resists. last point at bottom accelerates upward against gravity. 1G leading to Answer 2 ... If bungee stores enough energy to accelerate the man above 2 G's 1 G to overcome gravity. 1 G to equal same acceleration as falling. any extra energy would pull harder than falling. hardest right at the bottom. (only clean answer because it is only 1 point in time) true ... not enough info. without these assumptions. Thinking out loud. appreciate any correction

[@nolanjshettle]

To be safe I'm going to have to go with all of the above.

[@StormyMusic9]

Wait theres no option (f)? Where i can say the acceleration is the same at all points no?

[@waroftheworlds2008]

I feel like this would be a fun trig problem.

[@PiyushKumar-to7bd]

It is a problem of SHM as at its extreme ends

[@SrmthfgRockLee]

I've watched mythbusters and by guessing I think when you're most bottom for a brief second since then you kinda slow down n get back up.. My answer: "d."

[@flapjackattack4]

soo the acceleration of gravity is a constant... but no options?

[@markmongan8298]

I really enjoy your interactive vids! I chose option E :(

[@Lunara1990]

I said "B" because I misunderstood. I thought was when the jumper feels as if he's going the fastest.

[@vedprakashrai9118]

X=Asin(wt)....1 by taking double derivative we could easily get the result a=-Aw^2sin(wt)....2 this when wt=pie/2 at that point a is max and by putting wt=pie/2 in 1 we get X=A 😉

[@raman249]

Shm nahi hai ye!! It's true for small values of x......

[@gusanht7919]

The question is kinda tricky and it should be "When does he EXPERIENCE maximum acceleration?"

[@sanjayg6842]

4 varusham! How's life?

[@luckypegasusvol7700]

What is the actual answer?

[@karansinghmadhwal1074]

depend upou bulk modulous of the spring i think d because c point in spring is not streched but at d poin the spring is maxiummly streched and the force is equal to kx^2 x is lenth of steached spring if the object who jump in that bungey jumpie has lower wieght then #option_A is correct if that object is heavy then at d point acclaration is maximum

[@weirdoOo3s]

umm i think it's A because i believe that at that point the eye will most notice the change of speed and the the acceleration going from zero to whatever speed .

[@emacsjohn8357]

f/g = 1+2L/x at D. f>g. Ans is D.

[@reetanshukumar1865]

max acceleration should be on point where both gravity and tension parallel to each other ,but in this trajectory no such point exits.because tension always try to counter the gravitional force .so for max acceleration point we have to find where tension is zero .so max acceleration be "g" that is the point "A"..

[@dannyncorey]

Clearly 'C' .. considering the description of c was 'at his fastest point'

[@funicio]

acceleration, not speed

[@carajaneevans1785]

acceleration is different from speed. Speed is how fast something is, acceleration is how much the speed increases.

[@NetAndyCz]

a) the acceleration is g. b) it should start getting lower than g c) acceleration is 0 d) I am guessing d, though I think if the rope was elastic enough it could be lower than g, at least in theory. e) it is g again

[@Diedela]

There is no answer given? :(

[@danielkostovetsky9175]

At a, b and e he is in freefall, so his acceleration is approximately G Since the problem implies a point of maximum acceleration exists, we can eliminate all of these At c and d, rope tension is acting on him as well as gravity. The tension is greatest when the rope is tightest meaning d>c, so the answer is d (which is correct) We can alternatively eliminate a, b and e by inferring that the force of tension minus gravity at d is strong than gravity itself

[@connorrichardson5533]

Ive jumped off that bridge in queens town and im 12 I live in the north island :)

[@fesimco4339]

You're 17 now.

[@grundolord]

trick question, acceleration is constant... net force is the only variables

[@TheShadowCerberus]

The acceleration is greatest when you jump feet first and the cord flips you.

[@things4you540]

A i guess because from nothing going to moving is more than from fast to faster

[@BholeChaturee]

At A v-u/t is greatest as initially its was.. u=0 then it gained speed say 2m/s the the acc. is 2m/s2 but but afterwards v-u will reduce

[@aeonian_harshul]

Acceleration in free fall is 10 m/sec2

[@harrison_ingram]

D change of speed and direction👍🏻

[@LierreFrimaire]

Deep dark fantasies

[@harrison_ingram]

If the rope didn't stretch and he stopped instantly he would experience infinite g's and the universe would collapse on itself. Here is a thought. If you went from 0-1 mph instantaneously would you die? Yes. As slow as it is. Bc speed doesn't kill you. Acceleration does.

[@pleappleappleap]

That doesn't make the statement any less true.

[@1369ralph]

Isn't the acceleration the same everywhere? Like 9.8 m/s^2 Accept for E

[@1369ralph]

Oh wait, nevermind

[@toastbrot2153]

It's b, because there is the largest force action on the jumper

[@toastbrot2153]

Oh wait, then is also a and e. So my answer is a, b and e, because there is the gravitational force acting on the jumper. and this force is in every of the five cases the same. But in c and d, there is additionnally the force of the rope acting in the other direction on the jumper. Therefore, the total force is lower in c and d. And because force and acceleration are proportional (F=m•a), the acceleration is largest at a, b and e.

[@harrison_ingram]

Not e

[@harrison_ingram]

+toastbrot he is experiencing close to 0 g's

[@dandong8329]

answer is d. f=ma , mass constnat, bigger force, bigger a

[@kalexomarriba8814]

Ooh come on i am using a cellphone!!!

[@JavaLu]

Isn't the Acceleration the same where ever the jumper is???? I think this might be a trick question for Speed VS Acceleration. Anyway, lets pick a video and find out! =D

[@luckypegasusvol7700]

I just feel like a cuz, imagine free falling after some point you are gonna reach your max speed and stop accelerating... bruh not gonna lie i don't know

[@CaJoel]

RIP annotations

[@quasa0]

I've got the links for you

[@keith7357]

I'm using mobil device so u can't click on the screen

[@convincinglies1]

im gonna have to go with E charles

[@sauravamatya2755]

E ....like pendulum

[@lartedelvoloahazzodicane.3488]

Assuming a point mass jumper and weightless bungee the second newton's Law reads ma - mg - kx=0. so a= g + kx/m so the maximum acceleration occurs when when bungee is at its maximum lenght. so the answer is D.

[@bass-dc9175]

Easy: A and D because there is allways a balance of valocity and acceleration. Meaning the acceleration is the highest when he has 0 valocity. Edit:I watched the solution and have to confess that i misstook his term of acceleration: he didnt ment a.result so only D is correct.

[@bass-dc9175]

"Veritasium well said. Just to be sure I counted the frames till the rope was taut, and then counted the frames from there to the bottom. There are more frames accelerating than decelerating which also supports the conclusion that the magnitude of acceleration is greater than g at the bottom." So a.D > a.A meaning a.D > every other a (a for acceleration)

[@dixi3333dixi]

The answer is b

[@muhammadazeemashraf3007]

it is eithr a or they all are the same

[@gavinvales8928]

b, obviously

[@vahidmirkhani]

BEHOLD!!! The answer is "d" no matter what! Qualitative proof: at "d", velocity is zero but the guy starts to go up. Meaning the acceleration is upwards, so greater than 9.8. If it WERE 9.8, the guy would stay at the bottom with velocity zero, but he's coming back up, so acceleration at "d" prevails 9.8 and gets the guy back up. Quantitative answer: What's the actual proof? We know in this particular case you can approximately say that energy is conserved. If you take the initial height "h", unstretched rope length "L" and stretched length"X" added to L, you suddenly say (h=L+X). Now conservation of energy: mgh=1/2*KX^2 Remeber there's no velocity at "d" so all gravitational energy has turned to spring's potential energy. Solve the simple quadratic equation and you get: X = mg/K * ( 1 + sqrt (1+4LK/2mg) ) The sqrt is always bigger than 1. So KX = mg ( 1 + blahblahblah). So KX is ALWAYS larger than mg. There you go. Thumb this up so that other people can see the final answer too (^_^) You're welcome.

[@danielrhouck]

I'm fairly sure you solved the quadratic equation incorrectly. You used ones where you should have used zeros. Remember, if ax^2 + bx + c = 0, we have x = (-b ± sqrt(b^2 - 4ac))/2a. In this case, putting mgh = KX^2/2 into this form gives (K/2mgh)X^2 - 1 = 0, so X = sqrt(2K/mgh)/(K/mgh) = sqrt(mgh/2K) (we ignore the negative square root because we know X is positive). Even if your solution were correct, it would not prove D was the right answer. It proves that the gross accelation upwards is higher than the gross acceleration downwards, which we can see from the video without calculation. It doesn't tell us that the net acceleration is greater than the acceleration downwards due to gravity. As an example, you proved that at the bottom the acceleration from the bungee cord is greater than 9.8 m/s^2. Suppose it was 9.9 m/s^2. Then we have a net acceleration of 0.1 m/s^2 upwards, which is well less than that at a. We need to prove that the force from the spring is more than *twice* that of gravity, which is stated in the description for D but I don't think is true for all types of springs.

[@vahidmirkhani]

Daniel Houck I'm happy that someone has taken the time to read my answer (^_^) Thank you (^_~) Now back to our question: Dan, you're forgetting about the unstretched rope length which is a constant (I named it L) where in this video is case "b" (length L below "a" as I assumed we know, and can be measured from the video if you have the dimensions) minus case "a" (zero height). Why is this "L" important? B/c it tells you exactly how much "X" will be. Can I ask you something? Please click on choices a. , b. , c. and d. and listen to how Derek explains each choice, even if you know them already. Then go back to my solution and see how L is crucial in finding X. F = - K * X & I proved that this F is larger than m*g no matter what. But you already knew that! It takes a long time from choice c. (where m*g = K*X) to choice d. (where K*X is much larger). Finally, you are right!!! The last net force might be 0.1 m/s^2 upwards, or it can be even more than 13 m/s^2!! It depends on your initial height (in this particular case, your unstretched rope length L, plays that role!) which is crucial in my final answer. The problem is that I don't think he (Derek) is asking about the net force! Because these 2 forces are acting on 2 different spots: one is acting on the guy's feet and the other is acting on his center of mass. So his body will feel a stretch. This "stretch" is at its maximum at d. Yet the "net force" acting on his body as a whole can be dependent on L. Therefor none of the choices are necessarily right. It can be either a. or d., only one of these 2. So I agree to your answer where you said d. is not necessarily the right choice, but I think it does depend on the spring constant K, unstretched length L, and mass M. Please read my reasoning carefully and tell me whether I have missed anything or not (^_~) Once again, I partly agree with you, I just think your reasoning is not complete. Thank you Dan for reading this :D

[@KiraleosAkis]

Gravity is not a force

[@ibrahimayna2642]

You need to watch UMUT ÖNCÜL