Advance Illustrations - A Bullet Firing Toward a Balloon | Kinematics #1 for JEE Advanced

In this video Ashish Arora Sir has taken up a case of A Bullet Firing Toward a Balloon.
Watch Advanced Illustrations Kinematics videos on various concepts:
1 A Bullet Firing Toward a Balloon- • Advance Illustrations ...
2 Non-Uniform Motion of a Particle- • Advance Illustrations ...
3 Catching The Train at a Crossing- • Advance Illustrations ...
4 Bouncing of a Ball on Staircase- • Advance Illustrations ...
5 Stones Reaching Ground Together- • Advance Illustrations ...
6 Measuring Height of a Building- • Advance Illustrations ...
7 Time Required to Avoid Collision of a Car- • Advance Illustrations ...
8 Calculation of a Tower Height- • Advance Illustrations ...
9 A Bird Flies Along the Train- • Advance Illustrations ...
10 Acceleration and Retardation in a Journey- • Advance Illustrations ...
11 Ball Caught by a Fielder in a Cricket Match- • Advance Illustrations ...
12 Distance and Displacement by v - t Curve- • Advance Illustrations ...
13 Hailstones Rebounding off From a Windscreen- • Advance Illustrations-...
14 A Fighter Plane Hit by Anti Aircraft Gun- • Advance Illustrations ...
15 Time Varying Position Vector of a Particle in Motion- • Adv Illustrations -Tim...
16 Rectilinear Motion with Deceleration- • Advance Illustrations ...
17 Maximum Time in Reaching Ground- • Advance Illustrations ...
18 A Projectile Crossing Two Given Points- • Advance Illustrations ...
19 A Projectile Grazing Over a Triangle- • Advance Illustrations ...
20 A Cycle Crossing Busses Periodically- • Advance Illustrations ...
21 A Stone Hitting a Flying Bird- • Advance Illustrations ...
22 A Ball Rolling off a Stairway- • Advance Illustrations ...
23 Maximum Range of Projectile From a Tower- • Advance Illustrations ...
24 A Ball Clearing a Tower Projected from Another Tower- • Adv Illustrations - Ba...
25 Hitting a Berry in Shortest Time at a Given Speed- • Adv Illustrations - Hi...
26 Projection of Ball From a Truck- • Advance Illustrations ...
27 Elastic Collision with a Wall- • Advance Illustrations ...
28 A Swimmer Motorboat Following The Light House- • Adv Illustrations - A ...
29 A Swimmer Crossing a River- • Advance Illustrations ...
30 Throwing a Ball to Hit a Rocket- • Advance Illustrations ...
31 Projectile Between Inclined Planks- • Advance Illustrations ...
32 Shortest Distance Between Two Planes- • Advance Illustrations ...
33 Equal Components of Particle's Velocity- • Advance Illustrations ...
34 A Stone Hitting a Bird- • Advance Illustrations ...
35 Shots Fired From Top of a Hill- • Advance Illustrations ...
36 Distance Travelled in 2D Motion- • Advance Illustrations ...
37 Average Velocity in Projectile Motion- • Advance Illustrations ...
38 Perpendicular Motion of Two Projectiles- • Advance Illustrations ...
39 Chasing of One Particle by Another- • Advance Illustrations ...
40 Collision of Two Fired Shells- • Advance Illustrations ...
41 A Boat Crossing a River- • Advance Illustrations ...
42 Projectile in a Sliding Chamber- • Advance Illustrations ...
43 Mid-Air Collision of Two Projectiles- • Advance Illustrations ...
44 A Balloon Rising Above Earth- • Advance Illustrations ...
45 Particle's Velocity in a Parabolic Motion- • Advance Illustrations ...
46 Chasing a Cat by a Dog- • Advance Illustrations ...
47 Curvature of a Trajectory- • Advance Illustrations ...
48 Fluttering Flag Mounted on a Car- • Advance Illustrations ...
49 A Ball Clearing a Wall- • Advance Illustrations ...
50 Minimum Separation Between Two Projectiles- • Advance Illustrations ...
51 Distance Between Dropped Stones- • Advance Illustrations ...
52 Sliding Down an Inclined Plane- • Advance Illustrations ...
53 A Floating Raft and Boat Going Downstream- • Advance Illustrations ...
54 Separation of Events in Different Reference Frames- • Adv Illustrations - Se...
55 Flag on a Ship Fluttering Due to Wind- • Advance Illustrations ...
56 Cycling in Different Directions- • Advance Illustrations ...
57 Polar Trajectory of Boat- • Advance Illustrations ...
58 Firing a Shot on an Apple- • Advance Illustrations ...
59 Average Speed in a Rectilinear Motion- • Advance Illustrations ...
60 Projectile in Plane of an Inclined Plane- • Advance Illustrations ...
61 Range of Particle on Inclined Plane- • Advance Illustrations ...
62 One Plane Just Misses to Hit Another- • Illustrations-62: One ...
63 Maximum Height Attained by a Projectile- • Advance Illustrations ...
64 Two Perpendicular Trips in a River- • Advance Illustrations ...
65 Motion of a ball in Inclined Triangle- • 65. Advance Illustrati...
#jeeadvance #jeephysics #pgAdvIllustrations #pg4jeeadv #kinematics #rectilinearmotion #projectilemotion

[@Hardworkerpatshala123]

Simple apporch use In five seconds the ballon travels 125 meters and it has 25m/s upwards velocity v^2=u^2+2as for bullet U=0 bullet srts from rest a=10m/s s =125 On solving we get v=50 m/s Using relative velocity if bullet wants to hit the ballon the velocity must be 50+25=75m/s

[@bios546]

Sorry to bother you sir, but I just want to say... I've depended on this website for an year and it has taught me FIRST CLASS PHYSICS! because of all the questions, I'm able to solve tonnes of questions without being doubtful of the answers... Thanks to you, I've got the right amount of confidence... I seriously thank you for everything sir... Is there any way that people like me on this side of the screen can help the website? I reeeeeeally wanna help you make the website even better (it is really good now too!)

[@physicsgalaxyworld]

We welcome if you wish to contribute... pl write to ashish arora sir at his mail id [e-mail] in which way you want to contribute...

[@Sriram-sp7sl]

@Sai Krishna Prasad Sir can you share the expirence with physics galaxy.and also guide me such that i can handle jee advanced phyics...kindly if you have time let me knownthe answers plz...

[@ISHANISHARMAMV]

May you pls tell me...why we use relative velocity here in final height wala formula....(u-25) kyu likha h ..why not only u...pls clear this doubt.

[@richasingh9962]

@@ISHANISHARMAMV as we know formula of height is h= u^2/2g. We substituted the required velocity with u and solved it further we got the answer.

[@ISHANISHARMAMV]

@@richasingh9962 thanks

[@BinayakPanda1997]

What all things should i do from physics galaxy so that i can have an aim for full score in physics for jee main?

[@physicsgalaxyworld]

Go through all video lectures and Quset problems in sequence on www.physicsgalaxy.com as on youtube only video lectures are there qusets are not available...

[@BinayakPanda1997]

i dont have at all any time to watch advanced illustrations as because i am a repeater. will it affect so for jee main?

[@physicsgalaxyworld]

for jee mains not required... but practice is the main key factor...

[@BinayakPanda1997]

from where should i practice after quesets. In your book there a mix of mains and advanced level questions . I strictly want to practice only at mains level

[@mohanram7008]

@@physicsgalaxyworld Ok sir

[@Pw_insider07]

sir you have been working on the book for 9 yrs ....DAMN !!

[@mukeshsinghal9815]

Sir sorry to say but I do think this is wrong solution since in the interval bullet would take to reach towards the balloon should also be considered and hence the distance travelled by ballon should also be considered

[@Pentester5667]

Ye question too abj sir(Mohit tyagi wake)ne class mein bhi karwaya hai

[@maheshmangal8146]

A snapshot of a petrol engine is given in which piston is moving downwards with velocity 40sqrt(3)m/s.Find the angular velocity of the shaft : (A) 400rad/s (B) 300rad/s (C) 200rad/s (D) 500rad/s

[@vijayvyas5075]

sir their final relative velocities at the time of collision should be 25m/s and not 0. Am i wrong ??? please help me out.

[@ayushgupta-uf9ss]

Who is here after July 2022 for jee adv prep

[@THE___RISHU_]

Me...😐

[@JoyboyFistgk1]

​@@THE___RISHU_ hello mai bhi 2023 😅

[@THE___RISHU_]

@@JoyboyFistgk1 😀

[@rishabhjoshi7160]

Hmmm

[@AnimeNotion5865]

Drop

[@AnantMishra_anantdbest]

A body with mass 5 kg is acted upon by a force F = (-3î + 4ĵ) N. If its initial velocity at t = 0 is V = (6î - 12ĵ)ms-1, the time at which it will just have a velocity along the y-axis is ??? please sir give me the solution .. ans is 10 sec

[@physicsgalaxyworld]

+Anant Mishra Find vector acc = F/m and use its x component and x component of velocity to find the time after with Vx becomes zero.... So when Vx=0 then velocity left in body is Vy only... so we are doing 0=6-(3/5)t which is giving you t=10s....

[@nitish9684]

By apply kinematics 1st eqn in x axis put vf zero getting 10 ans

[@glory1950]

accl = (-3i + 4j)/5 Integration with respect to Time. v vector = (-3ti + 4tj)/5 + c Here c = vel at t = 0, i.e. 6i -12j So, vel along x = -3t/5 + 6 = 0 t = 10 s

[@nivasmummadi80]

Sir I think it’s wrong. Ballon will not freeze when bullet is fired at t=5sec, so we should equate 25t not 125

[@anishka8713]

Sir used the concept of relative velocity so i think the dist he considered 125 was the rel seperation b/w the balloon and bullet. Please correct me if i'm wrong :))

[@emmvsingh]

JEE Adv. 2024 wale...❤?

[@Future_iitian_]

Me hu na jaanu

[@Pranay_9]

@@Future_iitian_ idhar bhi agaye nibba

[@chaman9537]

​@@Future_iitian_ kaisi chal rhi hai bhai prep ka

[@Future_iitian_]

@@chaman9537 26 ke baad hi puchna

[@chaman9537]

@@Future_iitian_ dropper ho kya aap

[@jinux8297]

Sir, By the time the bullet would reach the height of 125m, wouldn't the balloon rise up as it is said in the question that the balloon is rising up as a CONSTANT RATE

[@Ameydheple]

Sir used the concept of Relative Velocity. As the Balloon is rising up with 25m/s. The balloon is assumed at rest and bullet is given (U-25)m/s Balloon is moving upward, but relative to it velocity of balloon is taken. Hope you got it.

[@Ameydheple]

@@unknownrealms8452 Ya. I ordered ADV illustration book. But started solving from videos. Only 40 days left for Advanced now. All the best 👍

[@Ameydheple]

@@unknownrealms8452 don't worry about competition. Just focus on improving yourself 👍

[@Britannia06]

​@@Ameydheplehow was your exam?

[@Britannia06]

​@@Ameydhepledid u succeed?

[@DarkKnightLives]

I think the solution provided to this question is wrong. Nowhere in the question is it said that the balloon stops after 5 sec. Since the balloon is constantly ascending it will continue to ascend even after the bullet is fired. The minimum speed is then found by taking the final velocity of bullet to be zero as has been done but assuming that the balloon is continuously ascending. The answer will be higher than 75m/s. Thanks a lot...A good question.

[@rushib5197]

The balloon is ascending even after bullet had been fired.... Hence relative velocity u-25... And displacement of bullet is more than 125 But relative displacement will be 125 So ig the solution is crct

[@Mortals_o]

Someone Didn't study relative motion

[@AmanKumar-us9hr]

Bro go study relative motion. ( btw bcz we are observing motion with respect to ballon it will be at rest we will see the bullets the motiin only and the bullet's motion too would have been affected and modified due to ballon at rest.{in observation}.

[@nitish9684]

​@@rushib5197totally agree

[@rishabhkedia9304]

the correct ans should be around 80. as the bullet will also take some time to go up till balloon and height of balloon will increase.

[@vaidehibaranwal9245]

Relative velocity method

[@neehar6409]

@@vaidehibaranwal9245 can we not solve this without using relative velocity? pls reply

[@Kyacheesehotum]

​@@neehar6409 We can . Assume that at time t=t bullet reaches ballon (and at t=0, ballon is at height 5*25=125 m) . So calculate distance travelled by bullet in time t and equate it to distance travelled by ballon in time t . (Ballon goes h=125+25t. You can find bullets height by using u as initial velocity blah blah ) . Then a quadratic eqn forms . For the bullet to reach ballon it's necessary that time t is real number right? And for minimum, discriminant of the quadratic becomes zero . From there we can find a quadratic equation for u . Find th value of u . Answer comes 75

[@Kyacheesehotum]

Oh yes , time t is basically 5 seconds ater the start of baloon rising

[@CycleEnder]

@@Kyacheesehotum 2 years ago….whats the use now lol

[@hemalipuri246]

This is really the best!!

[@History-Geopolitics47]

Yes

[@deepakparashar794]

Anyone June 2023 adv.

[@user-tf6kn3yr7t]

Yep 🖐

[@Ameydheple]

Yes 🙌

[@user-dh9rj7yp4q]

Bro now you are in which college

[@shreyastiwari5699]

Yes brother

[@sarthakjaniPC]

Which college

[@BinayakPanda1997]

i am a dropper. I have only 7 months left. My PLAN for PHYSICS jee main is ONLY to sit in front of computer and watch physics galaxy basic videos and solve all quesets SINCERELY along with those videos present with it. i am NOT doing anything else and nothing. i am NOT reading any books becuase of less time. Will i be able to acheive full score in jee main? If not, tell me what the strategy should be? If possible, i want an answer from Ashish Arora Sir!

[@physicsgalaxyworld]

To score full in jee... reading books is must... to connect with Ashish Sir, you can mail him at [e-mail]

[@physicsgalaxyworld]

to connect with Ashish Sir pl mail him at [e-mail]

[@brothersofosman4447]

@@physicsgalaxyworld Are there video solutions of quesets too?

[@physicsgalaxyworld]

Yes quesets solutions are there which will be available after Oct'18

[@nobarakugisaki_._]

Yeah, Can share something about your preparation back then when you were preparing for Jee main. It's been 6yrs from now... Looking forward to get a reply from you,.. (A junior)

[@raj-rl3zo]

Solved in one go. 🎉

[@faheem4977]

this can be done by 1 equation you just have to use am gm

[@BinayakPanda1997]

how should a repeater or dropper should utilise physics galaxy to get its most benefit for physics for jee main within only a span of 7 months being left to complete whole sylabus?

[@physicsgalaxyworld]

7 months is more than sufficient time... but the only way is consistency... you need to be regular at least for one hour honestly on watching videos... excess time and lower time both will be harmful...

[@GoofyAhhLion2937]

Thank you sir for providing jee advance questions for free on yt

[@intellect8444]

🗿

[@GoofyAhhLion2937]

@@intellect8444 🗿?

[@harrykane1347]

Sir, why u didnt considered g for the balloon? that will make its velocity non uniform

[@rajatdeepsrivastava8304]

It is given balloon ascends at constant speed making its acceleration zero

[@vinayakvermaverma3672]

@@rajatdeepsrivastava8304 yes

[@HeeWhooRemaiins]

@@rajatdeepsrivastava8304 same doubt thanks a ton

[@nitish9684]

Sme doubt tha ye baat man ki ki constant ki vjay se acc =0 but diya kha hai neglected gravity external acc 0 hoga gravity always act

[@rishankgoyal1097]

sir I think some thing in your solution is wrong pls consider it logically- Time taken by bullet to attain max height if fired by initial speed 75 m/sec=u/g=7.5 sec Max. height attained by bullet if fired with initial speed 75m/sec=u^2/2g=281.25 m Height attained by balloon in (5+7.5) sec with constant speed 25m/sec= 25x12.5= 312.5 m How can a bullet which will go upto 281.25 m can hit a balloon at 312.5 m. it seems 100 % wrong please clarify mistake.

[@ether3531]

No in this assumption u r thinking that at the highest point of the bullet would be the meeting point but its not true the bullet would touch the balloon and the bullet speed will be lower than 25 m per sec

[@rishankgoyal1097]

@@ether3531 thanks biswas your answer really helped me alot so if we had to solve it without relative motion we have to use calculus actually putting du/dt =0

[@adityahegde8184]

@@rishankgoyal1097 please explain,I also have same like urs doubt,

[@mr.comicalbanda8260]

hey this is because in this two cases are possible when it is max height and when at the point of meeting their relative velocity is zero. for the first case V=80.9m/s and Hmax= 327.25 and for the second V= 75m/s and Hmax= 250m hence the minimum value is 75 m/s. Also in this cases are possible because velocity of balloon is constant, so their values V and be possible depending on the height from ground but these are limiting cases............

[@AjaySingh0829]

You are Awesome Sir

[@Roadto10x]

Is it wrong to assume that total distance traveled by the bullet will be 125 + 25t, where t is the time taken by the bullet to reach the balloon, then using S = ut + 1/2at^2 (replacing u with u = v - at) get the value of t and then find u? I am getting t = (5 + 5sqrt(5))/2

[@ramakrishnan4356]

me too getting the same answer

[@matrixate]

This guy is imposing that the balloon stops.

[@ash143aish]

Mee too having the same doubt ......because the balloon does not stops after 5 sec and it will continue to go up at 25 m/s. Answer will be something around 81m/s

[@jayendraarunachalam1031]

then the answer is around 81m/s

[@jayendraarunachalam1031]

@@ash143aish absolutely ... you are right......

[@sridharbajpai420]

sir. advanve me inme se ek concept 100%, aata h kya

[@physicsgalaxyworld]

All advance illustrations are very important so we suggest you to follow all in sequence...

[@AdityaKumar-os9mp]

Kon kon 2026 advance ki tayari kar rha hai

[@aditya00078]

Is all these adv illustration are from the ashish sir 700+ advance illustration book for jee mains and adv itself

[@debasishmodak66]

Yes

[@ramakrishnan4356]

sir a doubt after 5 sec bullet is fired.let the bullet hit the balloon after t sec so distance traveled by balloon is 125+25t=ut-1/gt^2 where u is vel of bullet also 0=u-gt if i solve like this i am getting u as 80.90169944

[@sarthakchauhan8386]

Yes, absolutely right the correct answer is 25((root5)+1)=80.901

[@adityahegde8184]

@@sarthakchauhan8386 has sir really done mistake??😔

[@sarthakchauhan8386]

Yes bro just think!

[@shlokvinodthakur]

Sir why you have not taken g for baloon height

[@MydusOfthegold]

It's ascending with const acceleration

[@virajbagal935]

i tried to solve it in very trivial way as follows: as we want the minimum speed of bullet, let it hit the balloon with zero velocity, it means time travelled by the bullet is u/g where u is the initial velocity of bullet wrt to ground. As distance travelled by balloon and bullet is same, 25*(u/g +5)=u*u/g. after solving we get u= 50 . But this is not the answer according to u ... whats going wrong here??

[@physicsgalaxyworld]

+Viraj Bagal distance travelled by the bullet is ut-(1/2)gt^2 not u*(u/g) as bullet is retarding at g....

[@mamtachawda6016]

Sir but can't we use the equation 125 +25u/g = (u)(u)/2g but ans. is 25(1+5^(1/2) coz 125 m + after 125 m how much dis Ballon will travel should be max height of bullet fired and equating this is giving above equation then why my and yr ans. is different

[@anubhavdebnath6504]

@@mamtachawda6016 finally, found someone who got the same result as me

[@harshsinha1873]

​@@mamtachawda6016 i also have the same doubt

[@pratiksrivastav]

​@@harshsinha1873did you got the ans to this Q ?as I am also struggling with the same doubt.

[@sankipanda5906]

Sir do advance illustration are helpful in neet,aiims.Or it is O.k to see only basic videos of chapters by you.

[@user-or6lr7sb6n]

Jinko dikkat a rhi hai note kre ki final velocitu relatve m 25 hogi so min velecity=25+25root5

[@radhamay]

*NEVER GIVE UP*

[@mr.comicalbanda8260]

sir i have a doubt that final speed of bullet is zero with respect to ground frame, since we are dealing with moving frame so shouldn't the final velocity of bullet with respect to moving frame be -25m/s as Velocity of bullet with respect to balloon = velocity of bullet w.r.t ground - velocity of balloon with respect to ground...... Please clarify the doubt sir please sir

[@user-or6lr7sb6n]

Same

[@mr.comicalbanda8260]

Yes, I think I understood, that for the final velocity of bullet to be zero with respect to the ground is not the case of minimum velocity of bullet...

[@user-or6lr7sb6n]

Sir ka ans galat hai final 25 hi hogi

[@namansingh5276]

​@@okayyy__ Agar meet krne se phle ground frame mai velocity zero hogyi bullet ki toh wo kbhi catch hi nhi krpayega balloon ko. Aur Agar max height nikal ke 80 wala answer aya toh usme phle hi meet kr chuka hoga balloon aur phir dusri bar meet krega jb Balloon ise krega

[@anubhabdey6360]

Sir sirf ek doubt hai... Balloon pe bhi toh 'g' lagega... And agar balloon pe g laga toh relative acceleration of bullet will become zero... Sir this is the only doubt I have. Looking forward to your reply sir. Thank you

[@deskarnator2554]

Bro balloon ka constant bol rkha hai question mai yaani g aur uska acceleration cancel hone ke baad bhi 25 ms^-1 se upar badh rha hai

[@idkjust2524]

Baloon pe 35m/s² ka acceleration hoga net 25m/s² constant hogya na💀 aur vaise 11 mahine phele ki baat thi mai ab kyu bta rha

[@ajay0077]

Starting at 11:03

[@ajay0077]

See my finishing at last lecture

[@sarthakchauhan8386]

Sir, by the time the bullet reaches the ballon, the balloon must have risen a few metres and the bullet will not hit the balloon, the correct answer to this problem is 80.9016...using a completely different and correct approach.

[@rishankgoyal1097]

you are correct bro sir have committed some mistake but iam unable to trace.

[@sarthakchauhan8386]

think basics you'll figure out the correct way and correct answer...

[@adityahegde8184]

@@sarthakchauhan8386 please help me in getting solutions 😔

[@maanu3808]

Sir have not considered g in finding height

[@maanu3808]

So..

[@adityavyas359]

Can we get the time at which ballon hits the bullet?

[@darkgamer3738]

Bhai agar hm u relatve le rhe hai to g relative kyu nhi.

[@jeequick3]

U Variable hai g constant hai

[@ayushthelegend]

🔥Stop scrolling, go study🔥

[@subhashreebehera7761]

Sir when we calculate the height of the balloon then g is not considered why

[@nirmalya_nanda]

Sir I did it in a different way I formed a quadratic equation in t and for min u b^2-4ac=0

[@HangingQueen]

And use e=mc²

[@pushkarpallav8814]

sir why we are not considering here accln of bullet wrt baloon as we are considering relative velocity of bullet wrt ballon is u-25 then accln[bullet-ballon] should be -g-(-g) which would be 0 then why u have considered accln sir please gelp if i am wrong

[@jayendraarunachalam1031]

ballon is moving with uniform velocity ( no acceleration)

[@vatsalsomvanshi5892]

@@jayendraarunachalam1031 bhai niche toh g lagega hi na

[@himanshupandey8871]

in every case if two bodies meet then is the final relative velocity zero?

[@LEGENDGAMING-qb6dk]

Sir plz app neet ke liye bhi aisa video banyae kya enn question neet ke level ke hai

[@awake4586]

Sir can we do this question by taking the 125 +x distance covered by balloon in 5+t seconds and taking the same amount of distance for the bullet to cover in time t so that it can hit the balloon

[@Walnut_758]

I approached the same way but it is getting complicated like that there are 2 variables generated

[@CycleEnder]

@@Walnut_758ha same but time can be eliminated by using v = u - gt where v can be zero since minimum speed is asked

[@Walnut_758]

@@CycleEnder yeah but relative velocity saves like 2-3 steps that's why I said it gets complicated and not unsolvable

[@CycleEnder]

@@Walnut_758 true. Although I’m making some mistake and not able to match the answer using the longer method 🤦🏻‍♂️. Can you confirm if possible?

[@Walnut_758]

@@CycleEnder sry bro i tried but didn't got the answer here's the method: The balloon travels 25t+125 and time 't' the Bullet travel the same distance So, 25t+125=Ut-(1/2)gt² By substituting the value of t as u mentioned We get to the equation U²-50U-2500=0 By differentiation we try to find maxima and minima we get U=25 as minima which is in correct

[@abhinay5942]

U = 50 ms-1 w.r.t ground

[@gautammittal1481]

Who is here after September JEE 2020

[@aiming9011]

Me😁😁

[@SumitKumar-wy2px]

Me

[@vikrantshinde3051]

what rank didi you get in jee adv 2020

[@adityamishraiitbhubneshwar8039]

Sir i have done this question many time by other method but my ans is not coming out to be correct.... I have done... Let ball reaches its H max in time t1=v/g And max H = v^2/2g Then in this time t1 baloon is at a height =125+ 25t1 Then 1 equate both heights Can you spot my mistake sir plz

[@NotThatEnigma]

Unfortunately, sir have done a mistake here. Sir has taken relative velocity at collision to be zero which is not. Your method is absolutely correct, I was struck at this question for one day, finally figured it out.

[@NotThatEnigma]

Also, we can't use v^2 /2g formula for height for relative velocity.

[@rohanbarde5958]

Are bro, koi mistake nahi hai na sir ke solutions mein, na tere solution process mein!! But directly velocity nikalne se pehle, time nikal aur positive time quantity ko hi use kr solve krne mein!!

[@dakshjethva1490]

Sir in this...relative gravity wont be considered....which is zero...??

[@collegeonly8841]

Only bullet is affected by the gravity so we'll consider relative accn as g

[@piyushmourya3183]

Sir why are u taking relative velocity of bullet

[@piyushmourya3183]

Yaah! I got it......

[@abhinav6063]

@Physics Galaxy Sir, I'm able to do all the problems in the practice ex. in your book except for 2-3 questions (for which I had to get a hint from solution) which is lowering my confidence Am I doing it the right way or Should I change my strategy.

[@piyushgautam2943]

Bhai ho gya kya clear jee tera

[@PrinceGT]

​@@piyushgautam2943what about you

[@himanshupandey8871]

Sir can we use this approach of solving the problem : the bullet will have to cover the distance of 125 m in 5 s so using s = ut-(1/2)gt^2 and putting the values u =(u-25),t=5 s i am getting the same answer i.e. 75 m/s sir please clear my doubt

[@raman249]

the bullet is shot AFTER 5 seconds [note-the ballon is still going up which is why we took relative velocity u-25 i.e we are considering that ballon is at rest while the bullet travels with a speed of u-25 m/s[concept of frame and relative velocity]] what u said that bullet has to cover 125 m distance in 5 sec is wrong and we dont know the time it took to hit the ballon but using the formula S=v^2/2g we can find minimum velocity without knowing the time it took for the bullet to reach the ballon. EDIT-after 3months i replied to this post xd.

[@neehar6409]

@@raman249 so the only way of solving this problem is using relative motion?..

[@sujaynaidu598]

I did the same way and i got some 78 m/s

[@tsamrawat5448]

However you get correct answer but it is coincidence, method is correct but t is not equal to 5 sec, here t =(u-25)/g Using v = u+at V=0 ( final relative velocity) U= u-25 ( intial relative velocity) Luckily this time is also 5 sec ( note total time = 5+5=10), so u get the same answer

[@HridyanshUpadhyay-g2d]

I am a bit confused now please help me here sir has taken final velocity of bullet to be zero as we have to find the min initial velocity but ABJ sir explained this q without using relative velocity in which he has taken final velocity to be 25 m/s plz reply if somebody could help

[@matrixate]

I've deleted my comment because it's clear that you meant this problem for a beginner level student. I am not. So, make sure next time you upload a video, to consider that not only basic students will watch but graduates as well. I hope you do realize that my last comment that I deleted was in fact true about the aforementioned I deleted, which I won't mention here so as to not "confuse students" even though a couple of people here already see the problem with this example.

[@krzy_8340]

Wtf u taking about?

[@sachinparashar6640]

Sir we have to do these question in sequence or not, the level of the question is incresed one by one or these are of same level

[@physicsgalaxyworld]

You should watch these question in sequence first... all the questions of a topics... once you complete then after couple of hrs or next day try to attempt all these questions without looking at solutions and you'll realize that some questions you will not be able to solve despite of you've watched all solutions earlier... then rewatch the solution of those questions again to give impulse to your brain for your mistakes...

[@History-Geopolitics47]

@@physicsgalaxyworld 😍

[@harshpatelcse45]

Hello sir , i have a doubdt regarding books I recently discovered about pg channel and i already have brought allen's leader batch study material . So shall i buy the physics galaxy books or shall i solve the sheets ? Well the lectures are very helpful and benfecial thanku for all the efforts sir❤️

[@sejalsangole2527]

Just solve the sheets..

[@harshpatelcse45]

@@sejalsangole2527 i dont have any sheets

[@sejalsangole2527]

@@harshpatelcse45 I mean the Allen study material you're talking about

[@harshpatelcse45]

@@sejalsangole2527 ok i will try that

[@vishwajeetchoudhary8264]

while calculating H,why u didnt consider acceleration

[@jithendra.visualsz]

Because it is moving constant speed so acceleration 0

[@arjunsinghbisht9009]

sir how many percent of the questions do you give from I.e irodov

[@physicsgalaxyworld]

In advance illustrations we cover all variety of questions required for JEE Advanced whether these are from IRODOV or from other sources...

[@anusha6034]

@@physicsgalaxyworld Nice job sir Thanks a lot

[@003belekushagraupadhayay5]

i mean to say that can i download whole playlist in one go

[@devanshurajput3841]

Your voice in this video seems so unenergtic .. Maybe you were suffering from fever that time. But still u were making videos for us thank u sir..

[@vaibhavmourya65]

bro sir has voice like this only lmaooooo

[@virupakshkumbar7313]

I have completed syllabus for jee main , does watching advanced illustrations only complete all advanced syllabus and topics

[@keshavmittal4638]

Sir today I got the book Adv illustration 700 How to use the book efficiently As I have not enough time to watch the lectures? 2024 aspirant

[@utsavbalhara4618]

I am dropper, for jee 2024. 11th nahi padha tha. bro even 5 months are more than enough srsly

[@slingerv.p2611]

Sir ans is coorect but method seems to be wrong. U should write accn of bullet wrt balloon=-10m/sec

[@neerajkumargupta3220]

i got this question right !

[@singhsahab3642]

Sir you are best teacher like that Hc verma sir

[@kunaldas9543]

Itna asan hota he jee adv level question ❤ ... Baki ch ke vi esa hi he kya

[@ahdanish2732]

Sir if I complete all jee advanced illustrations will I crack jee advance exam easily plz reply me sir I am waiting

[@vroq2]

no its just illustrations we will need to do a lot of ques involving multiple concepts . these videos will help us to have all advanced concepts revised so we can apply them in the ques

[@ahdanish2732]

@@vroq2 OK but crack toh Ho hi Jaya gaa

[@ahdanish2732]

@@divyanshutiwari2150 ha

[@ahdanish2732]

But nit ma

[@ahdanish2732]

@@divyanshutiwari2150 ha concepts sikh lo

[@silvershades....6968]

Sir,where can I get solutions for Qusets?

[@navinkrsinha8792]

sir if i am able to solve them by myself then do i have a chance to qualify jee advance plz reply sir

[@physicsgalaxyworld]

Yes. But follow these advance illustrations in the method explained in CAPS-1 video by Ashish Arora Sir...

[@sahilgoel2034]

Sir can you plz reply that when an object thrown from an constant accelerated body then object also have same acceleration as that of body after losing contact from body.

[@sahilrajeshgajbhiye4968]

ty sir

[@purushottam5417]

On calculating height of ballon why g is not assumed if not assumed than why assumed in case if bullet

[@Krishna.1829]

This is because there are different kinds of forces in a ballon that cancels mg and velocity becomes constant

[@Samarthwillibecomeiitian]

Can you explain me more elders

[@VinayKumar-nr8uv]

excuse me sir, where can I find solutions to Physics Galaxy volumes? Please, kindly give the information...

[@physicsgalaxyworld]

Physics Galaxy 2e books are with solutions of all practice exercises and unsolved objective conceptual and numerical questions...

[@anuradhaverma4350]

My answer is 25(1+√5)m/s. Sir please tell if it is correct?

[@MRCRAZYSAURABH]

Start for advanced

[@nterbeastro-slayers9528]

Sir are these advanced illustrations good enough for jee advance because I found this one really easy

[@morijin5568]

same but I will still solve all the mechanics plus current and electricity chaps. I saw a few magnetism questions and they were all of very good level. this specifically was a class 9th question lol.

[@Nitankunal]

​@@morijin5568 means

[@azeiiacessaro7940]

Sir are these same questions given in the Adv Illustration book?

[@anshik6507]

Haan exactly same diye hai

[@devendrayadav-yq7yl]

sir , u've used the eq 0-(v-25)^2=2(-g)H where H is the relative distance between balloon and bullet . but u've assumed the final relative velocity in that eq to be zero but the relative velocity ideally is non zero ie the bullet just hits the balloon and its velocity becomes zero(as minimum speed req) but the velocity of the balloon is still 25 m/s(relatve V bb=25-0) . then why we assumed zero??

[@physicsgalaxyworld]

We are solving the problem relatively not in ground frame...

[@devendrayadav-yq7yl]

+Physics Galaxy okay ! but i didn't understand why the final relative velocity becomes zero . Are we assuming that the balloon and bullet at the same velocity or both have zero velocity(which is not possible for balloon)

[@physicsgalaxyworld]

both at same velocity...

[@devendrayadav-yq7yl]

+Physics Galaxy okay ! if both have same velocity at that height then the answer we found is not minimum i.e the bullet will have 25m/s speed upwards(same as balloon) and bullet will continue until its velocity becomes zero because of retardation due to gravity . we should have assumed that final relative velocity of bullet wrt balloon to be -25m/s i.e the balloon is going up and bullet just after touching the balloon have zero velocity .

[@physicsgalaxyworld]

No as if bullet speed decreases below 25m/s then the distance between bullet and balloon will start increasing and bullet will never be able to hit the balloon... as balloon continue to ascend at 25m/s... so this is the min velocity required at which bullet just reaches the balloon at speed 25m/s....

[@Divyanandp]

Sir ki drawing ❤❤

[@neeleshmeena9570]

sir i am repiting this year for jee and i have score well marks in jee advance i am going to watch your all videos as well studying book can i get good marks in jee mains and advance from your lectures + my full efforts

[@neeleshmeena9570]

@@phantomgamer4159 bhai taaref toh nhi kr skta bss itna keh skta hun.. I am in IIT DHANBAD ECE branch

[@neeleshmeena9570]

@@phantomgamer4159 yes but main part is ki sirf jagah KZbin pr padne se kuch nhi hoga... Jitni jyada problem solv kroge tbhi benefit hoga.... Becoz In my first attempt I did that which I had not reapeated in my drop year

[@neeleshmeena9570]

@@phantomgamer4159 books ko prefer sirf theoritical ke liye..and some good examples Problem solving ke liye sirf previous year papper ki books.... And KZbin channels Ashish Arora sir,Mohit Tyagi sir for Maths chemistry not specific and random topic wise videos

[@shaankapoor9074]

@@neeleshmeena9570 bhaiya,can you guide if all these advanced illustrations were helpfull?

[@bharatiyainnovator3185]

Bhaiya, are this illustration enough for jee advanced ?

[@swamivivekanandanglovidhyalaya]

Ye to 1st equation se solve hojaega simply

[@akashpaila6094]

Sir where can I get these kind of questions

[@stevejustin670]

Sir's book Physics galaxy

[@finalgaming4194]

Sir... I want to know that... For JEE advance... Iss your all advance illustrations enough.. If i understand and remember them all and do pyq of adv of that chapter... Will it be enough

[@physicsgalaxyworld]

Adv illustrations are good enough but Solving PYQ will help in building your mind orientation for JEE Adv exam…

[@003belekushagraupadhayay5]

sir,can i download all these videos on my pc. if yes then tell me how

[@physicsgalaxyworld]

Yes by using youtube downloader...

[@ashutoshdixit2806]

Why you not take acceleration wrt balloon

[@matrixate]

The balloon is rising without acceleration, only constant speed. There was a force on the balloon initially greater than gravity but as it's ascending the force of lift is equal to the force of gravity so, it's now just velocity, hence no acceleration.

[@akankshaarya1987]

Sir kya ye neet k liyev helpful h ???.......plzz reply sir

[@ganeshraut3458]

love you sir

[@rashmibajaj1396]

I have doubt in the soln, it is written in the Q that at t=5s the bullet isfired, its not written that at t=5s the bullet hits the balloon, according to me, the soln has an error, please clarify..?!

[@collegeonly8841]

Solution doesn't have any error... It's perfect.... Also time isn't used the question is solved with relative motion... Pls don't mix

[@vishalgupta895]

sir how much part of jee is covered by cbse books

[@physicsgalaxyworld]

JEE Mains paper is fully made from NCERT Text books concepts...

[@ISHANISHARMAMV]

May anyone pls tell me...why we use relative velocity here in final height wala formula....(u-25) kyu likha h ..why not only u...pls clear this doubt.

[@nitish9684]

Relative hai na

[@J_Prince24]

Thanks sir

[@nitish3300]

Sir this advance illustration help in advance exam

[@nicename3896]

1:07 why not balloon is retarded by g??? Uske liye bhi to S=ut+1/2gt^2 hona chahiye Par usse 0 arah aisa kyu pls explain!!!!!

[@whitehat3062]

Because it's given that 25m/s is speed with which balloon is ascending that means balloon is having accn=0 it means accn of magnitude g is applied to the balloon via some mechanical instrument so that motion of balloon doesn't get affected by ext..force or accn.....

[@praveenelango9490]

Why is it u-25?

[@physicsgalaxyworld]

Praveen Elango Its the relative speed of bullet with respect to balloon...

[@somahota7278]

Who is Preparing for JEE 2025 ? Like Here❤

[@mrinmoybhagawati6359]

Sorry to bother you again but I am a 11 grade student preparing for medical entrance. Should I follow adv illustration chapterwise or after completing the syllabus of 11th.

[@neetuverma8490]

Balloon par bhi g lagega. I think