Projectile Motion (of an Object Fired at an Angle) - A Level Physics
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- เผยแพร่เมื่อ 4 ก.ค. 2024
- This video explains the projectile motion of an object fired at an angle for A Level Physics.
Indirect fire (provided by the finest regiment in the British Army - the Royal Artillery) is the example I used in this video. When you have a projectile fired at at an angle (it initially has both horizontal and vertical components of velocity) it will follow a parabolic path. This video shows how you can solve such problems using simple suvat equations.
Thanks for watching,
Lewis
This video is recommended for anyone studying A Level Physics in the following exam boards:
AQA
CIE
Edexcel
Edexcel IAL
Eduqas
IB
OCR A
OCR B
WJEC
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3 minutes of him nerding out on weaponry before he even gets to the physics
i love it how you have all these toy models to explain things. :)
+nickydak1d I couldn't get my hands on a real one for the video!
My teacher explained this for 2.5 hours straight, and your 10 minute videos explain it so much better. Thank you so much :)
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
You are awesome.... All my exams are based on you.. Thank you for your help...
+Ishan Gauli Thanks - I really enjoy making these videos
+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...
Ishan Gauli so how much did you get 👀
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!!
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
damn, ive been stuck on this topic for 2 weeks and you just made it 1000x more clearer and simpler. thanks
I’m watching this for the 100th time. And I have my AS boards in a week
Perfectly simplified a complex topic, thank you.
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 ;-)
You make this really simple and easy to understand. Thank you.
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! 😊
you're the best, this video has completely changed this topic for me! i now understand it haha, keep the amazing work up!!!
ur videos are so good and well explained holy cow thank you so much!
This was so useful, explained really clearly thank you 👍🏻
Very clear explanation. Thanks.
Lifesaver! Thank you!
This video was really cool and really epic
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 sol 300^2xsin(2x30) over 10 = 7794m
love the edits!
Great video mate 👌🏼
tysm! this really helped
"Fifty thousand people used to live in this city, now it's a ghost town. I've never seen anything like it."
good video!
Thank you so much for your help
You are most welcome
no way. thank you soooo much!!!
Amazing video thank you very much
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?
why have you considered upward as positive and what does this upward refer to
How to extract the angle of the projectile if the distance is known to me as well as the initial velocity
What would happen if the object was fired/thrown from a height of 1.80m? Would the answer change significantly?
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?
Why do we use sin for vertical component?
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?
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
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.
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
+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.
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
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?
Hello, do your year 13 videos support OCR?
+alexbevan007 Yes they do. There is everything for both the current and new specifications. th-cam.com/channels/kZ6jZF-9uxY86jKLU4Wqyw.html
hows the acc 0? could you explain
For a level, in all projectile motion questions we can assume the air resistance is absent?
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?
In AS level, usually it's said air resistance is negligible.
Yep we’re taught u ignore anything and everything we hv trouble with 👍🏻 (if only we could ignore questions in exams too... )
Where does 150 come from? i don't get how you decide the velocities for vertical motion. Please help.
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.
May i know why horizontal acceleration is zero? It is the same for all artilleries? thank you.
Pyaie Phyo For this I assume there was no air resistance, so no forces that slowed it down horizontally.
I need help pleaseeeeee! when is the value of g (9.81) positive and negative... pls I have exam and I am freaking out
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 :)
Why couldn’t we do 1150/tan(30) to get displacement
What do u do if you are given an angle
+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?
I was unable to understand physics but after watching this video i was unable to understand english 😢
The acceleration is only upwards in the second half and not the first
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.'
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
The data given is to one significant figure, If I'm not mistaken.
+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).
+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!
+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!
Guns and Physics
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.
+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.
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?
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
yes SOHCAHTOA was used as we have the hypotenuse (300m/s) and were trying to find the horizontal component (adjacent)
Why is g sometimes -9.81 and sometimes 9.81? When is it negative and when would it be positive?
Thanks
Oh nvm you just said why in ur video
Extremely late to the video lol but as the ball hits the ground, isnt the velocity negative? so, -259.81
are you an army lieutenant specialising in physics? it seems this is the case.
Mohammad Rafi I was an army officer, now I'm a physics teacher.
It felt illegal studying like this💀
hey man 300 is 1 sf, lost my bet to my teacher:(
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.
do you do maths? if so, maths helps to understand where certain equation come from
@@bhg582 I do yes! But thats just even more confusing lol
are you aware of uplearn making money off your videos by using them to teach paying students and schools