Projectile Motion - A Level Physics
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- Опубликовано: 29 сен 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
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.
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! :)
Thank you so much! Physics makes much more sense to me now.
Continues... That velocity will be 20m/s (given). And that is the (unchanging) horizontal velocity. So the initial velocity has a horizontal component of 20 m/s. So the actual initial velocity of the projectile is 20 / cos 60. (adjacent/hypoteneuse = cos 60). cos 60 = 0.5. So initial velocity is 40 m/s. Hope that makes sense - and I have got the maths right!.
Sir do these videos cover Calculus based Physic?
In the horizontal direction, would initial velocity be the same as the velocity?
Not necessarily. It will be the horizontal component of the velocity. If the angle of the projectile to the horizontal is alpha then the horiz velocity will be v cos alpha.
Cool Video
thank you
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
Cameron Calder - Try ruclips.net/video/2vsAY-tp9js/видео.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.
"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 🥲
29:10 RIP Harambe :'(
35:58
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
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.
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.
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
Thanks. There is a full set of my A Level Physics Revision videos. I have put the link in the description above.
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
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.
"So now we've got that H=vtsin(Θ)" Top ten saddest anime deaths
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!
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.
You, my friend, are a life-saver! Thanks for creating and compiling all these videos together!! :)
Many thanks. I'm delighted they have been of some help.
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
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.
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.
Velocity is speed but with direction. So i) cant be true. ii) can be true (eg rotating ball can have constant speed but changing velocity because it is constantly changing direction iii) can't be true because any acceleration must result in a change of velocity (either in terms of speed of direction or both). d) is I think wrong since displacement is distance and direction. So you cant have a greater displacement than distance traveled. But someone may spot something I haven't.
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.
This is a bit easier. Use the formulae F=ma and v**2 = u**2 - 2as
Use the second formula with the values of s given in the question. v=0 and u = 9.5. Calculate a in each case. Then use F=ma for each value of a derived (m=20). If the force is 40 or more then the rope will break. So you want to find the value of a which results in the highest value of F below 40. Then the value of s which gave rise to the value of a is the shortest stopping distance.
Alas RUclips 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.
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
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.
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..
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)
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.
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 horizonatal axis - s (range) = vcosα t
For vertical axis v=u+at ie t=vsinα/g
So vcosα t = vcosα vsinα/g = v^2 cosα sinα/g
v^2/g is a constant. So range is greatest when cosα sinα is maximum
If x = cosα sinα then x will be max when dx/dα=0
dx/dα = cosα cosα + sinα (-sinα) = cosα^2 - sinα^2 = 0
So cosα^2=sinα^2 and cosα=sinα.
α=45 degrees
For horizonatal axis - s (range) = vcosα t
For vertical axis v=u+at ie t=vsinα/g
So vcosα t = vcosα vsinα/g = v^2 cosα sinα/g
v^2/g is a constant. So range is greatest when cosα sinα is maximum
If x = cosα sinα then x will be max when dx/dα=0
dx/dα = cosα cosα + sinα (-sinα) = cosα^2 - sinα^2 = 0
So cosα^2=sinα^2 and cosα=sinα.
α=45 degrees
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.
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.
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.
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).
I can't offer much advice on exam technique I'm afraid. My best advice is to make sure you understand the material as it is being taught, rather than trying to cram it in just before the exam. If you understand it then its much easier to get it marshalled inside your head. At least, that's what I found.
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.
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.
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.
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.
If the bullet is fired horizontally then it will never hit the monkey. The problem only works if the gun is fired at the monkey and the bullet travels at least the full range to reach the monkey. If it falls short then obviously the monkey is safe.
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).
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.
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.
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
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)
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.
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.
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?
They dont cover calculus as such. I assume people will have a working knowledge of calculus. But you might find some helpful material at Khan Academy.
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.
Yes. The monkey is safe if he is falling at a point beyond the range of the rifle. But if it falls within the trajectory of the bullet it has had it.
Well if you understand it now that's very good. But this is A Level material so dont feel too bad that it seems confusing. Stick with it. It gets easier.
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.
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.
I'm not sure what IB physics is. I cover the British A level exam system for students generally aged 17-18 ie just before they go to university.
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.
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.
Good question. I don't know. My guess would be that d might be confused with a differential eg acceleration = dv/dt.
Absolutely right. But not everyone would see that. The point of the calculation is to demonstrate that it is true.
You get a different solution if you use 9.81, but many exams use 10 to keep the calculations simpler.
ya physics is really a fun
It depends on your exam board. But there are certainly elements of it in some AS boards.
Quite right. It won't work if the bullet does not have enough velocity to get there.
My videos are intended to cover both AS and A2 courses in Edexcel, OCR and AQA.
What happens if the bullet landed before it reaches the monkey position in X-axis??? Where's the muzzel velocity of the bullet, should you take that into an account as well?
I use the up direction as positive and the down direction as negative.
It doesn't matter. It cancels out. 1/2m v1^2 = 1/2m v2^2 + mgh.
Looks like you've cracked it. Cos 30 is certainly squrt 3 / 2
Impulse (force x time) is change of momentum ie Ft = mv - mu
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.
animal activist dont worry the height was taken from monkeys hands so relatively bullet only pierced its hands. not a gr8 deal of damage!
a powerful teacher....i always recommend my fellows to use this channel
dude u r awesome. I understood all... thanks to u.
+MRA 7 a real lifesaver!
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 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.
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'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.
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.
I don't have a specific video on that, but you might find something of help in my video on momentum in two dimensions.
The key point here is that you always use the same convention for signs. In my case I have adopted the idea that the up direction is positive and down direction is negative.
Thanks for very kind words. Glad to have helped. Physics is fun.
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.
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
I have a video on vector addition, subtraction and multiplication which uses vector triangles to do addition and subtraction,
exam in 9 hours :) gotta love youtube for videos like this, + youtube lets you playback at 1.5X speed, and it's actually understandable at theat pace, soo woo more concentrated revision revision P.S to anyone reading this PIZZA IS LIKE OXYGEN, YOU CAN'T LIVE WITHOUT IT, and to anyone else up who has an exam soon, goood luck :3
Thanks. I hope the exam goes/went well.
It actually says 10 - 10t = 0. So 10t = 10. So t=1 sec. Hope that explains it, I used g=10m/sec^2 rather than 9.81.
Thank you soooooo much!!!!!! I finally get it now.....yassssss!!!!!!!! And I just subscribed too
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:}