Projectile Motion - A Level Physics
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- เผยแพร่เมื่อ 28 ส.ค. 2024
- A description of projectile motion, how a bullet or ball fired at an angle to the horizontal will travel through the air, and how to describe its trajectory. A full set of A Level Physics Revision videos is at • A Level Physics Revision
Very clear descriptions here that have really helped the students I teach. Thanks.
A Level Physics Online Thanks. Kind of you to say so.
Yo Physics Online stop stealing content off the real GOAT
when i looked at this tutorial... the thought came to my mind was...
"if i had such video 20 years back" learning would have been more easy and interesting.. i have always loved physics
:)
nice job..i hope kids of this generation make ample use of such great videos.
Cameron Calder - Try th-cam.com/video/2vsAY-tp9js/w-d-xo.html
may god bless u sir for making these videos..they really help a lot
I haven't done these calculations in 20 years. My son is now in AP Physics and needed my help. This was exactly what I needed. It is very well done and I am truly grateful for the refresher.
My Ib Physics test is comming up. All i can say is I love you for making this video
Thanks. I hope it went well.
I actually used this to help me with my NAPT test for the navy Thanks I needed a refresher
This is my first quarter of physics (in college and never took in high school) and I was really struggling to understand or follow anything that was going on. In this 30 minutes video, you cleared so many things up for me, I can't thank you enough!!!
How’s life now
I absolutely love the way you teach- it's all very easy to understand because you explain each step in the working out of a solution. You earned yourself a subscriber! Keep it up!
"And on that sad note, I hope that's been helpful." Haha, humorous and educative...thanks for the video!
I'm replying to your comment because I want you to look back in time
Damn, you commented 7 years ago, are getting any kind of nostalgia
@@A.K9488 And you commented 10 months ago, how're you feeling?
@@ranonymous27 Feeling terrified and depressed
Have my AS physics exam in 4 days 💔
Please pray for me 🥺🤧
@@A.K9488 I'll keep you in my prayers InshaAllah. Just wake up early, and utilize all the time you have revising and solving past papers, focus on the topics you find difficult and solve topicals for them, then leave it for God. Hope everything turns out well, I'm rooting for you, you've got this!!
@@ranonymous27 Thaankyou soo much for the kind words❤
You're soo nice
I'll try my best 🥲
I love the simple, clear no fluff teaching style. It reminds me of how I was taught back and school and later in sixth form. Thank you for your vids.
Yes. That is Newton's first law: a body continues in its state of rest or uniform motion unless acted on by an external force. So if no friction - and assuming the ground was flat - the ball would just continue to roll along the ground at uniform velocity since no forces would be acting in the horizontal direction.
For most questions on A-level physics projectile motion we make the assumption that there is no air drag or other forms of frictional resistance.
Thanks. There is a full set of my A Level Physics Revision videos. I have put the link in the description above.
I am a non physics specialist teaching A level physics for the first time this year. I'd be lost without your videos!!!! Thanks very much for such a great resource!
I'm very grateful for your kind feedback. I am also glad that the videos are being of some help.
As level people where are you at 🌚
Neeeeeeerrrrrrrrrd
@@imamgiuseppe5103 Geeeeeeeeeeeek 😪
So since 3 years passed what was your a level grade? 😁
Sir u r in which country
I've just chosen A level
hi, I'm a junior in high school, and physics is my hardest class right now, for a number of reasons, but mainly because I don't have a very good teacher. Thank you so much for posting this video, it helped clear up a lot of my confusion in projectile motion. You've accomplished in 36 minutes what my insufferable physics teacher couldn't in three weeks.
Well that teachers gone now😂
You, my friend, are a life-saver! Thanks for creating and compiling all these videos together!! :)
so today i had a quiz in my AP physics class and i didnt know anything about the topic we were on because i was in the hospital for almost all of it and i watch this in the morning and i saw my grade after i took and i got a 14/16 and thats without the curved so thanks to you i got a good grade because of how well you explained this. so thank you so much for the taking the time to upload this video :)
Thanks and well done.
no, thank YOU
the best youtube video you can ever find for projectory motion
I’m just starting on my engineering after a long years of gap, I find your videos very helpful. Thank you so much sir😊
a powerful teacher....i always recommend my fellows to use this channel
11 years later- still a legend.
As others have said, in the SUVAT equations each one misses one of the 5 terms. Generally, the question you are asked to solve will enable you to determine which one to use. You will be given some variables and asked to find another. Find the SUVAT equations which covers that group.
Best physics teacher ever. You explain everything so clearly.
I'm just starting out teaching AS Physics from GCSE, wow your videos are amazing! Clear, fully relevant and no nonsense.
Thankyou and I look forward to watching more of your videos as I progress.
Thank you so much for these videos! I am a student at high school in the Czech Republic and I am taking leaving exam in a few days. Without you, I would be lost. I did nothing for last 4 years.. and now, I found my textbooks so uninteresting. Your videos don't give me just information, but also motivation to study. :)
I would recommend this video to all my Physics classmates this clears up a lot of things, thank you very much :)
For Jake Cosgrove: This would be for example if the trajectory were to go up and then over a cliff so the projectile ends up lower than it started. The equations still work as long as your signs are consistent. Up = positive and down = negative. So for example the net vertical distance change is negative.
Hello. I want to search the approximate methods in quantum mechanics in particular the Hartree method, Hartree-fock method ab initio and density function theory . Possible to help me.
DrPhysicsA oo
This guy is super good at explaining physics. HELPED ME A LOT. Thank you
As the projectile describes the arc it has a horizontal and vertical component to its velocity. The horizontal component does not change. The vertical component gets smaller as the height above ground increases. So the tangent of the angle will decrease.
Got my exam tomorrow, this really helped to make me feel more confident
best physics youtuber by far
The ads kept on saying, 6 year old math problems doesn't help. Well, for me 90% of the things i learnt was because of the old videos. ^^ THANK YOU MR.!
Many thanks. I'm delighted they have been of some help.
This massively helped me tonnes, I'm resitting AS but I found your channel literally days before the exam! Gutted was an understatement. There's a question type bothering me at the moment with projectiles, where the horizontal range is extended by the height of the drop being longer on the second side of the parabola. If you or anyone else can help it would be immensely appreciated.
There are 3 parts to the question, a) and b) I have covered, it's c)
An Athlete is performing the long jump.
Horizontal Velocity= 8 M/S
Vertical Velocity= 2.8 M/S
a) Show that the total time the athlete spends in air is around 0.6s (assuming his centre of gravity is at the same height as it was at take off)
b) Calculate the horizontal distance jumped by the athlete.
c) In reality, when the athlete lands his centre of gravity is 50cm lower than its position at take off, calculate the extra horizontal distance this allows the athlete to jump.
Its been 8 yrs and no one answered😅
Wow this has been helping people out for years. I remember my physics 1 course looking at this vid.
Thank you so much. I learned more from this 36 minutes than I did in 6 hours of physics class per week.
29:10 RIP Harambe :'(
35:58
Not if you are talking about each being the same range (ie horizontal distance) from the source. The reason is that time taken is determined by the horizontal component of the velocity which doesn't change (since there is no horizontal force acts on it).
Thank you so much! You've allowed me to understand in 10 minutes what I've been struggling with at College for a month.
Thanks for very kind words. Glad to have helped. Physics is fun.
Thank you so much! Physics makes much more sense to me now.
Thank you so much for this lecture, it is much appreciated. Projectile motion used to be my weak point for mechanics 1 in physics, but this video has helped me grasp it and now in mathematics mechanics 1 it's possibly my favorite topic!
It depends what convention you choose. You are right. You could say that an object travels at 5m/sec in one direction and -5m/sec in the other. If you choose the positive direction to be upwards then remember that gravity will be a negative term (since it acts downwards).
Oh my gosh. i just wanted to take a minute to say thank you. i have just started to watch these because i have been really struggling with my physics. THANK YOU :') you really helped, thank you so much! ^_^ -Yanna from Cyprus! Subscriber gained:}
On your first point, this is the consequence of Newton's first law of motion, that an object will continue at rest or in uniform motion unless acted upon by an external force. Since there is no force in the horizontal direction, because gravity operates only in the vertical direction, there is no force to slow down the horizontal velocity. In practice of course, air resistance will slow an object, but we ignore that for these purposes.
Same principles apply. The trajectory will be the same except that it is 10m above the ground. If you need to calculate range then you have to calculate how far it will travel horizontally during the time it takes to travel up - reach a velocity of zero and then fall down to a position -10m below where it started.
No. The horizontal and vertical components are dealt with entirely separately. So if you drop an apple from a height of 1m above the ground, it will hit the ground at exactly the same time as a bullet fired horizontally from a gun (ignoring air resistance etc). In both cases the initial vertical velocity is zero.
For the range to be the same the two angles must add up to 90. So if one is say 70, the other must be 20. So sin α1 = cos α2.
The horizontal velocity = v cosα. The vertical velocity = v sinα.
The time to the max height =t. The time to complete the range = 2t (since time to go up = time to come down).
Use v**2 = u**2 - 2gs
If projectile were fired vertically then s= v**2/2g
For α1 0=(vsinα1)**2 - 2gs1
For α2 0=(vsinα2)**2 - 2gs2
s1 - s2 = 7/25 (v**2/2g)
Continued
My best compliment for the clarity of your tutorials. I've followed this video and it's everything crystal clear. You mentiioned that the distance is the same for all angle θ and (90 - θ). I stuggle to prove it mathmatically but I was unsuccessful. May you show me how to do so? Thank you very much, and thank you again for your video!!!
you are a man of letters in physics. I admire you sir.
This is an extremely helpful video! I'll surely go through this channel whenever I find difficulties in my A levels course. Thank you!
You are fantastic. This is helping me pass my physics subject at Sydney university. Many thanks.
Do you mean when a projectile is moving over an inclined plane, so for example a ball is thrown into the air but lands on the slope with a particular angle to the horizontal?
I am really thankful to you for giving lecture free of cost.It developed my concepts on projectile motion.Sir your lectures makes it an ideal foundation for those students who intend to take their studies to a more advanced level.Physics is my favourite subject.KEEP IT UP!
Sorry I didn't cover that but we can work it out from the info you have been given. The initial velocity of the projectile consists of a vertical component y and a horizontal component x. The horizontal component will not change. The vertical component will slow down (as a result of gravity) until it stops going upwards and starts to fall downwards. At the highest point it is traveling at it's minimum velocity (since its horizontal velocity is unchanged and its vertical velocity =0). Continued..
Thanks for the kind comments. Stick with it. It's a great subject.
The Coefficient of restitution e = v2/v1 where v2 is velocity after bounce and v1 is velocity before bounce.
Consider vertical component of velocity.
Use v = u + at u = 0 at top of bounce. So time taken to go from ground to top of bounce is t = v/a. The total time for ball to rise up and fall back down again is 2t = 2v/a.
The horizontal range = unchanging horizontal component of velocity (h) x time
R1 = h x t1
R2 = h x t2
R1/R2 = t1/t2 = v1/v2 = 1/e
You just earned yourself a new subscriber!
I'm not sure which graph you are referring to but in general you identify the two variables one of which is dependent on the other.
5 years late, but this has helped me so much. THANK YOU SIR
The components of the initial velocity (8 m/s) appear to be 8 cos 20 in the horizontal direction and 8 sin 20 in the vertical (down) direction. The horizontal component is not affected by gravity. But the vertical component is.
A huge lifesaver, thank you so much for taking the time in making this video!
Horizontal velocities dont change. Vertical velocity of both is given by v**2 = gt after t secs (where assume g=10). So combined velocity cmpts of 1st (x) is given by Pythag x**2 = 1 + 100t**2. Combined velocity compts of 2nd (y) is given by Pythag y**2 = 16 + 100t**2. Now when the two are at right angles x**2 + y**2 = square of combined horiz velocity = 25
So 1 + 100t**2 + 16 + 100t**2 = 25
200t**2 = 8
t**2 = 0.04
t = 0.2
Horiz dist = 5 * 0.2 = 1 m.
No, the term is correct. At any point on the trajectory the projectile will have a velocity (z) which has a horizontal (x) and vertical (y) component. The velocity z = square root of (x squared + y squared) - that's pythagoras. Now the horizontal component is the same throughout. So z will be a minimum when y = 0. Another way to look at it, is that at its highest point the projectile has maximum potential energy (mgh) and lowest kinetic energy (1/2 m v squared). Hence v is minimum at highest pt.
G is just the gravitational constant. potential energy is usually denoted as negative such that at infinite distances it achieves maximum value of zero. But that is just a convention. You can adopt any convention you like as long as you are consistent.
You can choose any signing convention as long as you are consistent. I take the "up" direction as positive and the "down" direction as negative.
the poor monkey sacrificed himself in the name of physics, but did not die in vain, for we all learned projectile motion from his death. what if the projectiles mass was insufficient to penetrate the monkey? then he'd have a red spot underneath his hair. all joking aside this was a very excellent presentation of ballistics in general. I loved the step by step processes our astute professor led us on. well done, thank you for not taking your viewers knowledge for granted and omitting necessary steps as some of our instruction is presented these days at least in my experience sometimes. I am a faithful subscriber sir.
In the diagram I have, perhaps confusingly, called the initial velocity of the bullet from the gun - v. That can be resolved into its two component parts. Perhaps it would have been better if I had called the initial velocity u.
I'm less sure about that. I suspect I cover most of the material for any A level syllabus, but if there are particular areas of main physics (as opposed to biophysics) that you think are missing please let me know.
The same principles apply. The key is to determine which direction is +ve and which is -ve. I usually use up as +ve and down as -ve. Go gravity is always -g. If a body is projected downwards then the vertical component of its velocity is negative.
H is the height above the point from which the gun was fired, if there were no gravity acting. h is the height above the ground from which the gun was fired, when gravity is acting. The point is that the hunter being unaware of the effect of gravity thinks his bullet will end up a distance H above the starting position whereas in fact it will only reach a position h above the starting position.
Thank you so much for ur time Sir. I might not fail physics now.
+Dark Sky Hope the exam goes well.
+DrPhysicsA
Are you still going to post more videos?
Isn't it a class 9th topic. Cause it is so easy
@@mannyheffley9551 mate, I was learning how to expand brackets in year 9. This is more A level based.
Are you given the angle at which the body is projected? For example if the body were projected vertically up its momentum when projected = mv = 2x20. When it reaches ground again its momentum = mv = 2 x -20. So change in momentum = 40 - -40 = 80. That's not an option.
This has got to be the best video for projectile motion! Thanks Sir
Well the same principles apply but are much more difficult to calculate. It depends on the motion of the ship. Basically the velocity of the projectile as measured on board the ship must be added to the velocity of the ship to get the position from the person observing from the shore.
Thank you soo much ! :) I understood more in 10 mins of this video than 2 hours in class. :)
It doesn't matter which convention you adopt as long as you are consistent. I choose the upward direction to be positive. So in my convention, the initial velocity is positive because it is upwards but gravity is always minus because it always acts downwards.
Not sure which theta you want to calculate but if you know the vertical and horizontal components then vert/horiz = tan theta.
This sounds like a pendulum question. If pendulum expands then its period of oscillation will change since period is related to length. You need to calculate the period of the initial length, then the period of the revised length. You can then work out how many oscillations there would be in 1 day at the initial length and by how much that would change at the new length.
This is one of the best videos I've found on projectile motion. Great job and thanks.
On your second point we have to be consistent about signs. I am using the up direction as positive and the down direction as negative. So gravity will always be negative in the down direction.
Yes. I do not separate the two because the playlist attempts to cover material required for a number of different exam boards including AQA, OCR, Edexcel, CIE
Thanks, this kind of explanation is what I wanted to hear for a long time
Amazing video, much better than reading a textbook.
You are AMAZINNGG!! You have cleared all my doubts regarding Projectile Motion. My Physics teacher should watch your videos and learn how to teach. You are a life saver!
Thank You so much!
May God bless You! :)
I hope so. My videos cover the bulk of the syllabus of most of the A-level physics courses. But at present they only cover a small amount of the physics that would be taught at university. Thanks for your comments and good luck for the future.
Now you can teach yourself every thing online. Amazing computer revolution!
I don't follow a specific Board - but the A level videos on this channel are intended to cover broadly the topics in an A Level Physics syllabus (both AS and A2) in the UK.
the video is specifically clearand i understood the equations and the methods of it. i gradually thank you
This man is the ultimate plug...
Thank you for clarifying all of the formulas!!
ur r great sir i have been back of this chapter for a long time but nothing was going to mind but ur teaching helped me gain back intrest in physics
Not sure which formula you are referring to but if all you have is an initial velocity (and not the angle) you cant really calculate anything.
Oh my goodness, what I can say is that they have not been of "some" help, but rather A LOT OF HELP! Thank you, thank you so much! I finally understand this topic! =D
Well I would say b) Statement A could be heating a gas. Heat is energy in (the cause) and the gas expands creating work out (the effect). Statement B. Energy is a scalar; it has no direction. Work is force x distance. Force is certainly a vector. Distance is a scalar on the basis that the direction is already specified by the Force.
My A Level videos are not based on a single exam board. They are a mixture of AQA A&B and OCR A&B but I haven't (so far) done the biophysics aspects.
Your explanation was clearly explained and included all the required details. Thanks very much.
Well if I've understood your graph correctly it suggests that the body is moving at a speed of 1cm/sec and is then reflected back at 1cm/sec. Change in velocity = 0.02m/s. So impulse = mass x change in velocity = 0.04 x 0.02 = 8 * 10**-4
"So now we've got that H=vtsin(Θ)" Top ten saddest anime deaths
Much much better than my physics teacher...thank u very much sir for your valuable video