Projectile Motion (at an Angle from a height) - A Level Physics
ฝัง
- เผยแพร่เมื่อ 18 พ.ย. 2017
- This video explains projectile motion of an object that is fired at an angle from a height above the ground for A Level Physics.
This is a rather long example with a projectile fired at an angle from a raised platform. I show you how to break up the motion of the projectile into three sections then use suvat equations for both the horizontal and vertical components of velocity.
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
_____________________________________
MY PHYSICS WEBSITES
Find even more videos organised by exam board and topic at:
GCSE Physics Online
► www.gcsephysicsonline.com
A Level Physics Online
► www.alevelphysicsonline.com
MY TH-cam CHANNEL
Your support in watching this video has been invaluable! To contribute towards the free videos on TH-cam, make a small donation at:
► www.paypal.me/physicsonline
FOLLOW ME
► th-cam.com/users/physicsonline...
► / physicsonline
► / gcseandalevelphysicson...
#projectiles #alevelphysics #physicsonline
![CAMPปลิ้น | EP.77[2/2] "รับทราบ" แก๊งเพื่อนรัก 10 ปี ที่ไปไหนก็มีตำนาน!](http://i.ytimg.com/vi/ifc0ofD51B0/mqdefault.jpg)
you the greatest teacher ever
This better be for 20 marks! 😂
Nah, it'll be 3 marks max.
i dont mean to be off topic but does anyone know of a trick to get back into an Instagram account?
I somehow lost the account password. I would love any assistance you can give me
@Harry Nathan instablaster ;)
@Jonathan Miller Thanks for your reply. I got to the site on google and I'm in the hacking process atm.
I see it takes quite some time so I will get back to you later when my account password hopefully is recovered.
@Jonathan Miller it did the trick and I actually got access to my account again. I'm so happy!
Thank you so much you saved my ass :D
Great help, thank you
thankyou. this was super helpful
why did you put a = -9.81 for part B and C, Should it not be a = 9.81, as now arrow moves down. i e. direction of gravity.
THANK YOU!
SUPERB EXPLAINATION SIr. @
Good luck everyone!
Wouldn’t the initial vertical be equal to u sin feta rather than it being the horizontal velocity
How is the initial velocity same for horizontal and vertical component? Shouldn't the vertical initial velocity be found by 100sin45
I was thinking the same thing. In the previous video, which practically uses the same principles, he did use sin for the vertical component of velocity.
In fact 100cos(45) and 100sin(45) are the same value.
@@worstbuffs9189 its the question, usually they arent the same
Hi why is 'a' negative for all three sections, when in section A it is travelling upwards. You defined upwards as positive, yes, so it makes sense that a = -9.81 as gravity is acting down...BUT in section B and C the arrow is now heading towards the ground so surely a is now positive...please could you explain?
Whichever direction you choose as the positive direction, you keep it the same.
Acceleration's direction isn't defined in terms of the velocity's direction, but with respect to your coordinate system.
@@ishanr8697 Surely not because the results of each part only make sense when you use a = -9.81 throughout, else for example v=u+at where v = 0 and u = 70.71, trying to find t, the object is travelling upwards, using a positive a would give a negative time
What effects would air resistance have on the horizontal component or even on the vertical component?
Horizontal velocity would reduce due to air resistance, and the acceleration downwards would be slower too.
Why is the acceleration upwards negative in all the three sectors and where did u get that height of 100m for section C????
@Michael Gardner safe g
@Michael Gardner another late reply but acceleration due to gravity will always act downwards and in this case since he took upwards as positive, downward acceleration at all the sections will be negative
lets us look at this convention.. Acceleration due to gravity is positive if both force (gravity here) and displacement are in same direction. g = + 9.8m/s2 when a body is dropped. Example A body is dropped means initial velocity is zero. Imagine body takes 3 seconds to reach the ground. g = + 9.8m/s2 meant that after 1sec the speed becomes 9.8m/s, after two seconds it becomes 19.6m/s and after 3 seconds it becomes 29.4m/s. It means that it strikes the ground with a velocity of 29.4m/s. g = + 9.8m/s2, positive sign meant that the body increases it speed by 9.8m/s for every one second. Lets look at it in the reverse manner when you throw a body with a speed of 29.4m/s, it becomes 19.6m/s after one second, 9.8m/s after 2 seconds and zero after 3 seconds at its maximum height. Here g = -9.8m/s2 negative sign indicates the speed is decreasing
now how would you perhaps find the angle at the time of landing?
legend
Hi i love your videos. You did a very goodjob in explaining the topic than my teacher. The only thing im confuse is why have you concluded that displacement c is -100m?
Since displacement is a vector, we can measure its change (Delta) by doing:
(during flight in part C)
Final distance (ground level) - Initial distance (In air)
0m - 100m = -100m
@@pahcreates378 ahhh i see that makes sense thank you for the help and sorry for the late reply 🙏🙏🙏
@@maulanaakbar7510 npp 😉
6:17 Shouldn't the a=9.81 instead of - 9.81 cos this time it's actually moving in direction of gravity?
No because +ve is upwards and it should stay negative if -ve is downwards
Doesn't matter whether the object is aligned in the direction of gravity. Gravity always acts downwards, it pulls you towards the earth no matter what part of earth your at
Bruh imagine if time was vector
@@archiebrew8184 hahaha that would be crazy
Hello Sir. Just wanted to ask why is s = -100m in part C, the only ''100 value'' I understand is the launched velocity at 100 ms-1 which was given. Please can anyone help me on this? Thanks :-)
oh nevermind I realised that was a given value by him at 6:24. It makes sense now.
I had the same question.
But the value shouldve been written from the beggining as the height of the cliff
Where has the -100m come from on part C? I don't understand as there is no distances labeled on the diagram. Thank you
he must have just made it up as a displacement valcue to use
He stated at the beginning of the video that the cliff is 100 m high
thanks@@eyiramcofie2504
at 6:08 why is the acceleration at B still negative 9.81 shouldnt it be positive since its going downwards??
NO our convention was +ve upwards
I was just wondering why is the time for vertical and horizontal components always the same?
the time in this is how long the arrow is in the air for, since time isn't a measurement of distance or speed, they don't affect it
7.58 sir how did you get -83.44. When I did -70.71^2x2(-9.81)x-100 I got -3038. Square root wouldn't work because its a minus number.
Check your working again. -70.71 squared is 5000. 2 x -9.81 x -100 = 1962. So v squared is equal to +6962. V = 83.44
Why can’t you use S=ut + 1/2at^2
Is it because it forms quadratic and that’s not allowed in Physics ( I do mechanics in A level maths and so we use this equation & S= vt- 1/2at^2)
@@Suraj-so3ge This is possible, you will need to solve the quadratic and pick the relevant root.
a bit late to ask a question but can I also find the total displacement by finding the displacement of each part and adding up? or would that not make sense coz you would have to do horizontal and vertical components separately?
yeah i still dont get it
because like for your final answer of total displacement, the time is just for part C? so like shoudlnt you add it all together?
oh wait no, he adds all the time together so it would make sense
@@kujojotaro3035 yeah i think all you need to do is add the displacements for the horizontal if you want the total horizontal displace ment aka range and for total vertical displacement i think its 0 since it hits the ground but im not sure. the only reason he adds time up is because its the same for both
@@stephenabraham2448 wait when?
is the vertical displacement at the end of B zero?
i would say yes
Yes because from the start of A to the end of B the displacement would be a straight horizontal line so there would be no vertical displacement so 0
hi,how did S become 100m?
how did the displacement become -100 i didnt understand that part please
apparently he said the the height of the cliff was 100m at the start of the video in which it's negative because it's moving down
3:24 but isnt u= 100sin45= 85.09??
Aeroplanes Lover No it isn’t. Because it is at 45 degrees you have the same component of velocity in the vertical and horizontal directions. Cos 45 = sin 45.
A Level Physics Online oh sorry my calculator was on radians mode.
Thanks a lot for videos. I understand physics more as I watch more of your videos.
Aeroplanes Lover Not a problem- a common mistake to make.