Projectiles and the Suvat equation : Introduction : Tutorial 1 : ExamSolutions

[@ASLUHLUHC3]

If anyone is confused (like I initially was) about when to consider vertical or horizontal motion, here's a guide. When finding Range or Maximum Height, it's pretty clear which direction of S (displacement) is required to solve for them: use vertical for Max Height and horizontal for Range. However, the required direction when finding Time of Flight is not so obvious. Consider horizontally if you have enough horizontal information about the system to use t=s/u. This simple equation (essentially from speed=distance/time) can be used in the horizontal as there is no acceleration, which eliminates the need for suvat. Otherwise, consider vertical motion using suvat. Hope this helps

[@LoneWolfMidnight]

You are a living legend. You may have potentially saved my physics A-Level. Honestly I feel in debt to you and this is stealing. If only my school teachers were as good as this!

[@ExamSolutions_Maths]

Thanks and I wish you all the best

[@kingrobert7246]

fr why the physics teachers always gotta be so bad

[@doseoffaiyaz]

​@@kingrobert7246 Because they hire them through whether they have certain academic qualifications or not. Not by whether the person truly has the ability to teach. I would rather go with a teacher who did bad in life in terms of grades and might not have the best of academic qualifications but truly has the ability to teach in terms of his/her field of expertise.

[@sharmisthatripathy9623]

I am a perfect example of not scoring or bad in maths in high school but now I am one of the best maths tutor for A level in my area.

[@pcmediaexpress901]

This mans voice is so nice to listen to, my entire Year 12 with my maths teacher shoving on these videos and sitting back, and I wouldn't have it any other way lol

[@ExamSolutions_Maths]

When taking the horizontal motion a=0 so all the suvat equations reduce to s=ut or s=vt since u=v The speed is constant horizontally. It never slows down or speeds up horizontally so v does not equal 0 due to friction it just remains the same as the initial horizontal component of velocity.

[@ExamSolutions_Maths]

@GriffoPolitics Hi FP3 will follow after FP1 and FP2. Will be after June though.

[@rawrrii1020]

The way you saved my life wow😭!!!

[@GriffoPolitics]

I'm doing M2 (and possibly re-sitting M1) in June, these videos are absolutely fantastic for me. Are you planning on doing some FP3 tutorials? I have that in June, too, and I need a top grade to get into Uni. Thanks.

[@ExamSolutions_Maths]

Because it is at the same horizontal level as the initial velocity but going in the opposite direction.

[@ExamSolutions_Maths]

You used the wrong value of T. You need to halve it for the max height, then it should work.

[@AkashManick-h6r]

Impressive. Do u also give teacher course in maths and physics

[@aashav2573]

When I used s = ut + 1/2at^2, for the time of flight, I get a different t value to you, I get 4.56 s instead? Thank you

[@rrrvna]

you were in radians not degrees. change it to degrees. (for anyone else) this comment is very old

[@Lau19K]

This was the most helpful video I've seen yet! Thank you!

[@georgemathew8218]

GOOD EXPLANATION Sir. @

[@thepanther1728]

how do we know its cos and sin on which side?

[@Fudgey1010]

Why is u not 0 in the first example? Obviously it isn't but why not? I got told that u is the starting velocity, which is 0 because the particle is not moving to begin with

[@ExamSolutions_Maths]

If it were dropped from rest then yes. However, if it was at rest to start with then you could not throw it upwards. If you had a ball and threw it upwards, you would need to give it a speed when it left your hand and you released it.

[@Fudgey1010]

ExamSolutions Ah ok that makes sense, thanks :)

[@skatertreflipable]

Hi can you help me with the first answer because I am getting something different even though I am using the exact same values

[@ChaksFM]

Ur calculator was probably in radians

[@joshyboy2210]

Hi, I've come to uni to study chemical engineering without having done mechanics or physics at A-level. I noticed your videos are very good at making things clear, in particular with equations of motion. Can you point me in the right direction to a running order for your video tutorials/lessons so I can learn everything I need to know for Mechanics (particularly equations of motion, SUVAT), from start to finish. I need to start with a basic understanding and progress onwards. Many thanks!

[@ExamSolutions_Maths]

www.examsolutions.net/maths-revision/syllabuses/Edexcel/period-1/M1/module.php

[@fickle_little_thing]

for range, there are 2 formulas apparently . one: if you know time: dx= vx / t or dx= [Vxsin2(theta)] / 9.81 its weird because sometimes it works but sometimes it doesn't. but on one of my exams i didn't have time and i used it but got .5 off because I had 22.5 instead of 23.5 or something. but the video i was following along with before i came here, (Projectile Motion by Tony DiMauro) , I calculated the range before i looked at how he did it with the second formula, but i got it wrong.

[@jcn268]

Explained it much more clearly than my teacher

[@fatmafaizal07]

hello, do you have a video on vector methods with projectiles?

[@ExamSolutions_Maths]

Not at the moment

[@isaacade2075]

sir at 7:46 why is the V not 0

[@carlaprata-xe9ws]

Thank you Sir!

[@ExamSolutions_Maths]

Thanks

[@_thekonrad]

ExamSolutions lol

[@andalahadel]

Why is g=-9.8 when you are calculating max height? Isn't the ball going up? acceleration due to gravity acts downwards so g= -9.8 however for max height the ball is going up so g=9.8, its the half motion.

[@ExamSolutions_Maths]

When an object falls freely to earth, assuming there is no air resistance, it accelerates at a constant rate of approximately 9.8 ms-2. Here I look at a stone being thrown upwards from the ground. Note: At 4:48 in the video I discuss the important point about the acceleration due to gravity, what sign to take and common misconceptions at 5:08. kzbin.info/www/bejne/qGeTZaSMm66ibNk

[@shubrotoshuvo867]

Hey do you have tutorial for chemistry and physic of CIE A2 level? If any user knows it plz suggest me one. I need it fast

[@halax3169]

YOU ARE MORE THAN PERFECT

[@ExamSolutions_Maths]

Tell my wife that!

[@samjab9237]

🛑hi sir, i wanted to thank you for the time you put into educating us and let you know i’ve subscribed! also, i had a question which i would be grateful if you could please answer. 😊 why is the V=-U in this video but in your second video V is unknown. thank you very much. 🛑

[@ExamSolutions_Maths]

At what point in the video is v= - u. Confused

[@aryaagunavante9898]

@@ExamSolutions_Maths I assume she is referring to 7:51 , where you mentioned that v=-30sin40 . Why is v = -30sin40 . Does the acceleration caused by gravity not affect the value of v? Thank you :)

[@lucahervada3697]

@@aryaagunavante9898 I am confused at this a bit too. However V is the speed just before the particle hits the ground, and since acceleration (-g) isn't changing, there should be symmetry. So initial velocity will be final velocity, but V will be negative as he took up to be positive.

[@flowerpower487]

@mynameistessa you need to halve the time.

[@mechamartini]

for the range , why can't we use the s=(U+V)/2 x T formulae for this ?

[@ExamSolutions_Maths]

You can but it is pointless as u = v for the horizontal velocity component so s = 2u/(2T) which reduces to s=ut. Which is what I used.

[@mechamartini]

Thanx , needed that.

[@aprilla9112]

thanksssssss

[@aprilla9112]

ur so cool

[@ExamSolutions_Maths]

That's okay

[@dawityohanes8280]

Zus gearit

[@adonis1up]

"Metres per second per second"

[@ASLUHLUHC3]

Lol he's not wrong

[@shahnazakter3907]

That’s a very long video can u please talk faster and do things faster pelase

[@ExamSolutions_Maths]

Nope but you play at 1.5x the speed or fast forward. You can't please everyone as some want it taken at a slower speed.

[@shahnazakter3907]

ExamSolutions sorry I was just not in a good mood. Ur video is perfect and detailed. I appreciate it😊

[@ExamSolutions_Maths]

No problems, I can confess to being in bad moods at times, it is just the way we handle it. Best wishes.