Projectile Motion - A Level Physics
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- Опубліковано 12 вер 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
Cameron Calder - Try ua-cam.com/video/2vsAY-tp9js/v-deo.html
may god bless u sir for making these videos..they really help a lot
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.
Thanks. There is a full set of my A Level Physics Revision videos. I have put the link in the description above.
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.
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
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
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!
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.
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.
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.
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.
"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 🥲
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
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😂
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
You, my friend, are a life-saver! Thanks for creating and compiling all these videos together!! :)
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.
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.
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
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).
29:10 RIP Harambe :'(
35:58
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.
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.
Many thanks. I'm delighted they have been of some help.
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.
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
a powerful teacher....i always recommend my fellows to use this channel
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).
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.
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.
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
I’m just starting on my engineering after a long years of gap, I find your videos very helpful. Thank you so much sir😊
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.
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.
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.
11 years later- still a legend.
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.
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.
the best youtube video you can ever find for projectory motion
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.
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:}
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.
Not sure which theta you want to calculate but if you know the vertical and horizontal components then vert/horiz = tan theta.
Thank you so much! You've allowed me to understand in 10 minutes what I've been struggling with at College for a month.
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..
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.
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. :)
This is one of the best videos I've found on projectile motion. Great job and thanks.
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.
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!
Thanks for the kind comments. Stick with it. It's a great subject.
Good question. I don't know. My guess would be that d might be confused with a differential eg acceleration = dv/dt.
My videos are intended to cover both AS and A2 courses in Edexcel, OCR and AQA.
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?
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.
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.
Alas UA-cam doesnt seem to let me continue with the answer. Maybe it doesnt like formulae. But work out s1 and s2 for each angle. Then s1 - s2 = 7/25 of the height above. If you replace sin alpha 1 with cos alpha 2 you should be able to get that
sinα1**2 - (1 - sinα1**2) = 7/25
sinα1**2 = 16/25
sinα1 = 4/5 so sinα2 = cosα1 = 3/5
Since ranges are equal vcosα1 * t1 = vcosα2 * t2
t1 / t2 = cosα2/cosα1 = cosα2/sinα2 = 3 / 5 / 4 / 5 = 3/4
If I have got the maths right.
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.
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.
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.
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.
I would recommend this video to all my Physics classmates this clears up a lot of things, thank you very much :)
Thanks for very kind words. Glad to have helped. Physics is fun.
Best physics teacher ever. You explain everything so clearly.
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
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😅
This guy is super good at explaining physics. HELPED ME A LOT. Thank you
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.
Can you tell me the time on the video where this arises please?
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.
You used v=u+at. I assume your vertical initial velocity is 7.07m/s. At the highest point v=0. So 0=7.07-9.8t. t=7.07/9.8 secs. So it takes 0.72 secs to get to its highest point.
You can decide your own choice of coordinate system as long as it is used consistently. I just chose to make up the positive direction.
Sounds like a conservation of energy question. So the initial kinetic energy must equal the final kinetic energy plus the potential energy. Kinetic energy is 1/2 m v^2 and potential energy is mgh.
There is a video on Geometric Optics - A Level Physics in the A Level Physics playlist.
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
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!!!
This is an extremely helpful video! I'll surely go through this channel whenever I find difficulties in my A levels course. Thank you!
Thank you so much! Physics makes much more sense to me now.
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!
This has got to be the best video for projectile motion! Thanks Sir
I don't follow any particular board's syllabus. I just try to cover the subjects which feature generally at A level in the hope that they may be of help for revision purposes.
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! :)
best physics youtuber by far
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.
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
You just earned yourself a new subscriber!
Wow this has been helping people out for years. I remember my physics 1 course looking at this vid.
A huge lifesaver, thank you so much for taking the time in making this video!
Thank you for clarifying all of the formulas!!
Well you can always reduce to a single equation. v**2 = u**2 - 2Fs/m. Only s is unknown. Of course the force is expressed in kg wt instead of the proper SI units of Newtons. So 40 kg wt = 40 * 9.81 Newtons. (9.81 is grav acceleration)
Many thanks for your kind comments.
The rifle is pointed at the position of the monkey before it starts to fall. Then as the monkey falls, the bullet also "falls" in the sense that it falls below the point at which it had been aimed. Without gravity the bullet would have reached the aimed target in t seconds. But with gravity it reaches a point x meters below that aimed point. Sadly, the monkey also falls x meters in t seconds. So the bullet and the monkey coincide. (ignore air drag etc)
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
Absolutely right. But not everyone would see that. The point of the calculation is to demonstrate that it is true.
the video is specifically clearand i understood the equations and the methods of it. i gradually thank you
"So now we've got that H=vtsin(Θ)" Top ten saddest anime deaths
dude u r awesome. I understood all... thanks to u.
+MRA 7 a real lifesaver!
You are fantastic. This is helping me pass my physics subject at Sydney university. Many thanks.