Projectile Motion (of an Object Fired at an Angle) - A Level Physics

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  • Опубліковано 21 гру 2024

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  • @georgiarobsonpowell5544
    @georgiarobsonpowell5544 2 роки тому +61

    3 minutes of him nerding out on weaponry before he even gets to the physics

    • @Phyron101.
      @Phyron101. 3 місяці тому +1

      meh it was funny thoug plus whos complsining over 3 mins

  • @sumayafelic1518
    @sumayafelic1518 4 роки тому +101

    My teacher explained this for 2.5 hours straight, and your 10 minute videos explain it so much better. Thank you so much :)

    • @beesechurger1
      @beesechurger1 10 місяців тому +3

      bit embarrassing really isn't it? quick 10 mins yt video is better than someone who is being paid and has 5 hours to teach us a week

  • @34293nbgauli
    @34293nbgauli 8 років тому +103

    You are awesome.... All my exams are based on you.. Thank you for your help...

    • @PhysicsOnline
      @PhysicsOnline  8 років тому +25

      +Ishan Gauli Thanks - I really enjoy making these videos

    • @34293nbgauli
      @34293nbgauli 8 років тому +81

      +A Level Physics Online You have absolutely no idea what your videos mean to student like us... On Behalf of many students like me... Thank you your noble effort...

    • @alkane2561
      @alkane2561 6 років тому +4

      Ishan Gauli so how much did you get 👀

  • @nickydak1d
    @nickydak1d 8 років тому +128

    i love it how you have all these toy models to explain things. :)

    • @PhysicsOnline
      @PhysicsOnline  8 років тому +118

      +nickydak1d I couldn't get my hands on a real one for the video!

  • @matthewbarker9188
    @matthewbarker9188 6 років тому +28

    You are the definition of a savior, I have mocks this week and all i needed was a slight recap, you are an inspirational human being and there needs to be more people like you

  • @lynx24100
    @lynx24100 7 років тому +83

    Love the military themed context to these videos!! Really helps to understand projectile motion and suvat and makes it way more fun. Also really interesting to know how this type of physics is applied to weaponry like artillery!!

  • @aaransarna9029
    @aaransarna9029 2 роки тому +8

    damn, ive been stuck on this topic for 2 weeks and you just made it 1000x more clearer and simpler. thanks

  • @AURAXKIRA
    @AURAXKIRA 9 місяців тому +3

    quick tip, at 3:52 you can find the horizontal range (s) with one quick formular this only works if the projectile motion is symetrical so no change in starting height and if air resistance is NOT taken into account. The formula is R = v^2 x sin2(theta) over g so 300^2xsin(2x30) over 10 = 7794m

    • @ABR-RLRook
      @ABR-RLRook 26 днів тому

      Oof that looks harder/ more confusing

  • @3nonti
    @3nonti 8 років тому +7

    Thanks a lot you almost saved my life I was like crying because I couldn't solve a question in my assignment, but now it seems I got it after all. keep it up . thumbs up ;-)

  • @mikailasurya
    @mikailasurya 2 роки тому +3

    you're the best, this video has completely changed this topic for me! i now understand it haha, keep the amazing work up!!!

  • @harin4965
    @harin4965 7 років тому +10

    I’m watching this for the 100th time. And I have my AS boards in a week

  • @matthewknight7616
    @matthewknight7616 2 роки тому +2

    Perfectly simplified a complex topic, thank you.

  • @ramithusellahewa3929
    @ramithusellahewa3929 2 роки тому +3

    Thank you so much sir for this amazing video:) I have my AS physics exams in about 4 hours and this really helped! Again, thank you so much! 😊

  • @gauthampaul4213
    @gauthampaul4213 4 роки тому +5

    ur videos are so good and well explained holy cow thank you so much!

  • @avi_mukesh
    @avi_mukesh 7 років тому +2

    You make this really simple and easy to understand. Thank you.

  • @kawsara5366
    @kawsara5366 4 роки тому +4

    "Fifty thousand people used to live in this city, now it's a ghost town. I've never seen anything like it."

  • @khizarhassan4152
    @khizarhassan4152 3 роки тому +3

    This video was really cool and really epic

  • @StanleyHung
    @StanleyHung Місяць тому

    Your physics teacher will launch artillery at you if you are still sleeping while the clock passes 7:00-8:00

  • @johnbiddleston8987
    @johnbiddleston8987 2 роки тому +1

    love the edits!

  • @lowkeycam9151
    @lowkeycam9151 Рік тому +2

    tysm! this really helped

  • @aa-qi2it
    @aa-qi2it 2 місяці тому

    This video was really helpful, thanks!

  • @physicstutus9851
    @physicstutus9851 8 років тому +1

    This was so useful, explained really clearly thank you 👍🏻

  • @studywithyan697
    @studywithyan697 Рік тому

    Hi, Why is this: 8:54 ? Can you explain it further to me please? Do we assume projectile motion as symmetrical at Alevel (OCR)?
    I have asked chat gpt about this and I feel a little confused.:
    Chat GPT:
    ' No, an upward projectile motion is not symmetrical. Projectile motion refers to the motion of an object that is launched into the air and moves along a curved path under the influence of gravity. In an upward projectile motion, the object is initially launched vertically and then follows a parabolic trajectory as it rises and falls back down.
    The trajectory of an upward projectile motion is not symmetrical because the time taken to reach the maximum height is different from the time taken to descend back to the initial level. This is due to the influence of gravity, which affects the object differently during the ascent and descent phases.
    During the ascent, the object slows down due to the opposing force of gravity until it reaches its peak height. On the descent, the object accelerates under the influence of gravity, increasing its velocity as it falls back down.
    Therefore, the upward projectile motion is asymmetric in terms of time, velocity, and displacement, as the ascent and descent phases are not symmetrical.'

    • @lowkeycam9151
      @lowkeycam9151 Рік тому

      ik this is kinda late, but for this topic, we assume that there's no air resistance or any other external force acting on the projectile so yea, projectile motion will be symmetrical

  • @Mirsab
    @Mirsab 5 років тому +3

    For a level, in all projectile motion questions we can assume the air resistance is absent?

    • @snhtsljedc9803
      @snhtsljedc9803 4 роки тому

      k DUDE!!! I know the equations but I struggle to use it atimes, like I I donno where to put it, do u have any advise?

    • @ayshakhatun00
      @ayshakhatun00 3 роки тому +2

      In AS level, usually it's said air resistance is negligible.

    • @Osamu_Dazaiii
      @Osamu_Dazaiii Рік тому

      Yep we’re taught u ignore anything and everything we hv trouble with 👍🏻 (if only we could ignore questions in exams too... )

  • @alamawy
    @alamawy Рік тому

    How to extract the angle of the projectile if the distance is known to me as well as the initial velocity

  • @Bino_Sleevelesssocks
    @Bino_Sleevelesssocks 7 років тому +2

    Why is g sometimes -9.81 and sometimes 9.81? When is it negative and when would it be positive?
    Thanks

  • @zarahasan6819
    @zarahasan6819 Рік тому

    hows the acc 0? could you explain

  • @syedarizvi7290
    @syedarizvi7290 7 років тому

    Why do we use sin for vertical component?

  • @Mirsab
    @Mirsab 5 років тому

    4:00 how do we find out u? Why did you use Cos 30? I'm probably supposed to know this but I don't, someone please help me!
    Did we use SOHCAHTOA?

    • @muhammadnumaan2798
      @muhammadnumaan2798 5 років тому

      its because the velocity that the projectile is fired at is 30 degrees were we need to find the horizonal velocity so we use cos

    • @cb818
      @cb818 3 роки тому

      yes SOHCAHTOA was used as we have the hypotenuse (300m/s) and were trying to find the horizontal component (adjacent)

  • @futureflash3479
    @futureflash3479 9 місяців тому +1

    Thank you so much for your help

  • @rubbaanrubbaan6454
    @rubbaanrubbaan6454 4 роки тому

    Great video mate 👌🏼

  • @jesscads123
    @jesscads123 6 років тому

    Lifesaver! Thank you!

  • @burmesedragon8851
    @burmesedragon8851 6 років тому +1

    May i know why horizontal acceleration is zero? It is the same for all artilleries? thank you.

    • @PhysicsOnline
      @PhysicsOnline  6 років тому

      Pyaie Phyo For this I assume there was no air resistance, so no forces that slowed it down horizontally.

  • @aryalalgonullu7150
    @aryalalgonullu7150 Рік тому

    What would happen if the object was fired/thrown from a height of 1.80m? Would the answer change significantly?

    • @six-winged-juni
      @six-winged-juni 23 дні тому

      If you mean “same flight path but starts 1.8m in the air” then I assume there would just be an extra bit at the end of the flight path, covering 1.8m vertically still at the same acceleration value (9.81). Would probably just be 1 or 2 more steps?

  • @skullrazor
    @skullrazor 8 років тому +1

    Very helpful videos, thank you.
    I just wanted to know how did u know when to use Cos (angle) x Displacement, or Sin, or Tan? Help

    • @BENS19777
      @BENS19777 7 років тому +1

      Skull Razor it's using soh cah toa. If you look at this example you're given the adjacent length and the angle to find the hypotenuse length.

  • @letslearnwithrahatfatima7336
    @letslearnwithrahatfatima7336 6 років тому +1

    why have you considered upward as positive and what does this upward refer to

    • @six-winged-juni
      @six-winged-juni 23 дні тому

      Guessing upward is considered positive because it’s the initial vertical direction the projectile is fired in? (Answering this mostly for anyone else who comes along later)

  • @mariehill6547
    @mariehill6547 4 роки тому

    If you were just using the vertical component of the object coming downwards, u would be zero, but would v be Usinα or would that be an unknown?

  • @kawsarmoqadsa5990
    @kawsarmoqadsa5990 7 років тому +1

    I need help pleaseeeeee! when is the value of g (9.81) positive and negative... pls I have exam and I am freaking out

    • @har0111890011
      @har0111890011 7 років тому +2

      Depends on what YOU decide is the positive or negative direction so if you chose up as positive then your g will be negative ofcourse as it is always acting downwards. Hope that helps :)

  • @gabrielhirst8711
    @gabrielhirst8711 8 років тому

    apparently one can find the maximum projectile displacement using a single equation : 1.225 * (change in time) ^2. Is this true and if not is there a different formula?

  • @BalaCliffDezjnr
    @BalaCliffDezjnr 2 роки тому

    So you’re assuming that if there’s no air resistance, the acceleration for x = 0 and since y is affected with gravity therefore y = 9.8 m/s2

  • @ModerateMeercat
    @ModerateMeercat 2 роки тому

    Why couldn’t we do 1150/tan(30) to get displacement

  • @josefalfarokreiner4549
    @josefalfarokreiner4549 7 років тому

    hello im writing my MAths IA and i will do it about kind of the same topic and i would like to know if there is a formula ?? no just formulas from physics?

  • @Mahmzo
    @Mahmzo 8 років тому +1

    Hey I have always wondered, How many sig fig shall I give? I don't wanna give the wrong amount. Is it usually 3sf? Or does it depend on the max num sf in the question

    • @PhysicsOnline
      @PhysicsOnline  8 років тому +8

      +Mahmzo Give your final answer to the least amount of significant figures quoted in your raw data. So if your raw data had values to 2 sf, 3sf and 4 sf then you could only justifiable give your final answer to 2 sf.

  • @Trickytriples
    @Trickytriples 8 років тому +1

    Hello, do your year 13 videos support OCR?

    • @PhysicsOnline
      @PhysicsOnline  8 років тому +1

      +alexbevan007 Yes they do. There is everything for both the current and new specifications. ua-cam.com/channels/kZ6jZF-9uxY86jKLU4Wqyw.html

  • @alexhall4002
    @alexhall4002 3 роки тому +1

    good video!

  • @ommerchant15
    @ommerchant15 7 років тому

    Where does 150 come from? i don't get how you decide the velocities for vertical motion. Please help.

    • @champion171299
      @champion171299 7 років тому +2

      Om Merchant Sin(theta) = Opposite/Hypotenuse
      Rearrange to give Opposite = Sin(theta) x Hypotenuse
      And the vertical motion is the opposite as it is opposite theta.

  • @akmedia8206
    @akmedia8206 2 роки тому +2

    Amazing video thank you very much

  • @ABR-RLRook
    @ABR-RLRook 26 днів тому

    8:26 you made a mistake here, you said 3972.6 and rounded it up to 3770 bc your 9 looked like a 7

  • @loothsattar3641
    @loothsattar3641 5 років тому

    How do we solve this if we don't want to find the maximum height. I feel as if we would get more information for the horizontal S or horizontal T if we just wanted to find where the bullet land or would we always have to split the motion of the projectile into two separate parts?

    • @six-winged-juni
      @six-winged-juni 23 дні тому

      The horizontal and vertical components (If i am correct about this) act separately from each other, so you have to calculate stuff for them separately. E.g max horizontal distance may use a different equation to max vertical distance, but you would still need to use SUVAT with the vertical velocity and stuff to find the time it’s in the air for.

  • @aronblemz1
    @aronblemz1 8 років тому

    no way. thank you soooo much!!!

  • @jackdevin387
    @jackdevin387 8 років тому +1

    The data given is to one significant figure, If I'm not mistaken.

    • @PhysicsOnline
      @PhysicsOnline  8 років тому +2

      +Jack Devin This is the danger when not using standard form. 300m/s could be to one, two or three significant figures and 30 degrees could be to one or two. But generally trailing zeros count as significant figures. In this videos I should only have given my final answers to 2 sig fig (but any intermediate steps should always be kept to the value calculated).

    • @jackdevin387
      @jackdevin387 8 років тому +2

      +A Level Physics Online I'm not sure if it's just my specification (AQA), but they explicitly state in their 'Practical handbook' (How fancy), that trailing zeros as you referred to them as, don't count as significant figures as that could be wrote as (3x10^2) - 1 sig fig, and my Maths exam board portray the same idea. Although, it could be argued that giving your answer to 3 sig fig was completely acceptable as the data book value of g was given to 3.
      The pedantic side of Physics, I suppose.
      On the note of Physics, anyway, your videos are brilliant and some of the best made I've seen. Your students are very lucky!

    • @PhysicsOnline
      @PhysicsOnline  8 років тому +4

      +Jack Devin I can see how a discussion like this opens up a whole can of worms. Stick with what your AQA exam board use. I should have given my data in standard form and this would have made it very clear the exact values used.
      You should only give your answer to the least amount of significant figure in the question, so even if 'g' is given as 9.81 the answer I should have given would be to two significant figures (provided the angle was actually to two significant figures as well). If you've been using Isaac Physics then you will be well practised in the correct number of significant fugures.
      Thanks for the comments on the videos - it has taken a great deal of time!

  • @JustBaileey
    @JustBaileey 7 місяців тому

    Extremely late to the video lol but as the ball hits the ground, isnt the velocity negative? so, -259.81

  • @jakey12369
    @jakey12369 8 років тому

    What do u do if you are given an angle

    • @PhysicsOnline
      @PhysicsOnline  8 років тому

      +jakey12369 Do you have an example? When you write down suvat for both the horizontal and vertical parts of the motion just make sure you calculate the initial velocities using the angle given. Does that help?

  • @AbdulRahman-vy7ko
    @AbdulRahman-vy7ko Рік тому

    The acceleration is only upwards in the second half and not the first

  • @AMIR55312
    @AMIR55312 8 років тому

    Is it like this that it is taught in the UK?, in my country we solve it with integrals. And it confuses the hell out of us XD.

    • @ALevelPhysicsOnlineYear13
      @ALevelPhysicsOnlineYear13 8 років тому +5

      +LeGunslinger This is how I teach it in the UK for A Level Physics. It can be solved with integrals but not every student in the UK also does A Level Maths and the Mechanics modules so they don't all know about integrals. This method works well - it takes a lot of practise though to really understand projectiles.

  • @f22rafi
    @f22rafi 7 років тому +1

    are you an army lieutenant specialising in physics? it seems this is the case.

    • @PhysicsOnline
      @PhysicsOnline  7 років тому +1

      Mohammad Rafi I was an army officer, now I'm a physics teacher.

  • @amshumanrajesh
    @amshumanrajesh 6 років тому +2

    Guns and Physics

  • @TopG20073
    @TopG20073 Рік тому

    It felt illegal studying like this💀

  • @abdultarar1408
    @abdultarar1408 2 роки тому +2

    I was unable to understand physics but after watching this video i was unable to understand english 😢

  • @Osamu_Dazaiii
    @Osamu_Dazaiii Рік тому

    A level physics is VERY confusing (maybe I shouldn’t hv picked it but oh well)... its like they teach u stuff and u cant question it. The more questions u ask the more complicated it gets. We’re soo brainwashed dat if my physics teacher pointed at a while table and said it was black, I wouldn’t question it.

    • @bhg582
      @bhg582 Рік тому

      do you do maths? if so, maths helps to understand where certain equation come from

    • @Osamu_Dazaiii
      @Osamu_Dazaiii Рік тому

      @@bhg582 I do yes! But thats just even more confusing lol

  • @vqlencjia8977
    @vqlencjia8977 8 місяців тому +1

    are you aware of uplearn making money off your videos by using them to teach paying students and schools

  • @kanishqnangia3657
    @kanishqnangia3657 2 роки тому

    hey man 300 is 1 sf, lost my bet to my teacher:(