I cant tell you how much this helped me its honestly amazing to see how much help you can get online and im so glad that a person like you has made these resources so easily available and simple to understand. Kudos to you!
I have a question regarding the question in 6:08, shouldn't in projectiles, the initial velocity be always equal to 0? Although, I see your point that at maximum height the velocity is equal to 0 as well.
unless there is air resistance the only force in projectile motion is purely vertical (the weight) and that causes acceleration in the vertical direction only and we use a=0 for the horizontal direction. Hope this helps!
@@zhelyo_physics Ah i see, I was confused cause you used -9.81 to calculate the horizontal range at the end but what you're doing is using the vertical motion to calculate the horizontal range. Thanks
for the course which I teach a calculator is allowed, otherwise you can use several approximation techniques or interesting calculation techniques. I am definitely not an expert on that though : ) Have a good research of common techniques if this is part of your exam.
yep! please use the syllabus as a tickbox though to ensure you cover everything. I think the only thing missing is the AC current, pretty much everything else is applicable.
nope, you are confusing it with the displacement equation. The units on the equation you mentioned are not right, and that's how it can be spotted (m/s)=(m/s)+ ms^(-2)*s^2 i.e. units on the left are not the same. Hope this helps!
This channel undeniably deserves + 1M subs! Crystal clear explanations with the most accurate notes! Thank you so much Zphysics, for all your help!
Thanks a lot for the comment! Much appreciated!!
Hi from Zimbabwe 🇿🇼 my brother I really appreciate your content. Keep on Making physics a fun subject to learn
Hey I'm also from zimbabwe
Thank you ❤
I've been watching your videos for a while now, and they've really been helping!
Great to hear this, thanks a lot for your comment and glad they are helpful!
This was super helpful, thanks!
Glad to hear this, no problem! : )
I cant tell you how much this helped me its honestly amazing to see how much help you can get online and im so glad that a person like you has made these resources so easily available and simple to understand. Kudos to you!
Fantastic to hear! Thank you so much for the comment!
Thanks so much! Amazing video.
anytime!
Amazing content, well explained. Thank you
No problem! Glad it has been useful!
perfect channel for me!
fantastic to hear, thanks!
You're saving lives bro 🙂 thank you
thank you very much for the comment! very much appreciated!
finally someone that makes me understand this thank you sooo much🙃
Anytime! Thanks for the comment!
Really amazing explanations. Thank you.
Thanks! Glad go hear!
thank you so so much!! quite helpful :))
Anytime! Thanks for the comment and glad this is helpful!
Thank u bro u are wonderful
thank you for the comment!
You saved my day .......thanks soooo much❤
anytime! thanks a lot for the comment!
Sir how would you know how many sig figs to make the answer? sometimes it's unclear
generally use the smallest number of sig figs given in the question.
Perfect
Thank you!
perfect!!!!!! Thank you so much sir
anytime!
sir can we only subscript Vy only when we subscript U with y as well?
or can we subscript Vx and Uy no right?
Generally we can choose whichever subscripts we like as long as it's clear and consistent. Hope this helps!
Thanks really helped.
anytime, thanks for the comment!
Thanks for this😭
Any time! : )
Thank u soooo much u really helped me alot
anytime! Glad to hear and thanks for the comment!
THANK YOU SO MUCH
anytime, thanks for the comment!
I have a question regarding the question in 6:08, shouldn't in projectiles, the initial velocity be always equal to 0? Although, I see your point that at maximum height the velocity is equal to 0 as well.
Nope we take the initial velocity after the force has been applied. Hope this helps!
Thank you, sir.
anytime! thanks for the comment!
amazing video
thanks a lot!
Late question but when do we use a=0
Ive seen in many cases where in the horizontal direction a is given as 0 in the suvat
unless there is air resistance the only force in projectile motion is purely vertical (the weight) and that causes acceleration in the vertical direction only and we use a=0 for the horizontal direction. Hope this helps!
@@zhelyo_physics Ah i see, I was confused cause you used -9.81 to calculate the horizontal range at the end but what you're doing is using the vertical motion to calculate the horizontal range. Thanks
is this ocr A aswell
Definitely, this is the spec I actually teach in college
@@zhelyo_physics ok cool because i just started a level physics and u helped my gcse a lot
But what if there is air resistance?
I have an entire video for this funnily: th-cam.com/video/9KvmYiCpZBU/w-d-xo.htmlsi=JnXISK9oiltz9OKY
Awesome
thanks for the comment!
Thank you sir
anytime! Glad to hear!
Hey this is a good video but how do I calculate the sin(35 degrees) without using a caluclator
for the course which I teach a calculator is allowed, otherwise you can use several approximation techniques or interesting calculation techniques. I am definitely not an expert on that though : ) Have a good research of common techniques if this is part of your exam.
Sir is this content applicable for Edexcel International AS level?
absolutely. Projectile motion is in all exam boards.
thanks man
Anytime, thanks for your comment!
Can we get the notes of this lecture?
sorry only the videos are available
hello, can you share the notes that you made in these videos?
it'll be great help!
sorry I have not actually kept them when filming, so only the videos are available.
What is use to make this video Please🤩🤩
Hi, this was made with Microsoft whiteboard, which is free.
are these videos applicable to caie exam board
yep! please use the syllabus as a tickbox though to ensure you cover everything. I think the only thing missing is the AC current, pretty much everything else is applicable.
Thank youu so muchhh@@zhelyo_physics
isn't Vy=Uy+.5at^2
nope, you are confusing it with the displacement equation. The units on the equation you mentioned are not right, and that's how it can be spotted (m/s)=(m/s)+ ms^(-2)*s^2 i.e. units on the left are not the same. Hope this helps!
@@zhelyo_physics yh ,it did ,thank you very much
Why is the time 1.17s instead of simply 117s....how can 58s give 1.17s. Awesome explanation though.
Which part of the video exactly? the time of drop is 10sin(35)/9.81 = 0.58 so times 2 it gives 1.17.