I never write comments so when I do you know I mean this so sincerely. Physics has never ever made sense to me since middle and high school. Now in college I gotta get it so I don't fail these classes. This is the first time I've felt confident and actually understood even slightly what I was doing. So thank you Derek!!
Wow the quality and organization of his calculations as well as pace and verbal description of the problem beats even KHANA. Great quality content I'm subbed
Absolutely AMAZING! I can't get my Physics teacher to explain any of this to me and my book is too confusing to help me with these kind of problems. Thank you so much for doing this video! I understand it so much more now.
Yes, that's correct. That's because for a projectile, there is no acceleration horizontally, so the velocity is the same the entire time, and you can find it with the equation x/t.
Thank you so much Im doing my first year of biomed but i didnt learn much about physics in high school. I know theres projectile formulas but I've been trying to find videos where they use the kinematics one since that increases your understanding. Thank you so much for making these, it was very clear!
I typically use v_0 for the initial velocity. That's v with a zero in the subscript position. It basically means "velocity at time zero" or "velocity when t=0". Then I use v by itself for velocity at some later time. Using u works also, though. Good notation definitely helps, but it's not as important as the concept.
The key here is to distinguish between the vertical and the horizontal. Once the bomb is released, it is a projectile, and there is no horizontal acceleration. There is vertical acceleration, though: gravity is pulling it down. It also gets more complicated when you include air resistance. Air resistance introduces some horizontal and vertical forces, which depend on the speed.
@Ell4Sh Yes, that approach would also work. There are typically multiple ways to set up a problem, all of which should give the same final answer. I set it up with 0 a the top because all of the motion and the acceleration are downward, in this problem, and setting it up this way makes all the displacement, velocity, and acceleration numbers positive. The other approach is fine, though.
One of the key points of projectile motion is that the horizontal motion is independent from the vertical motion. There is no acceleration horizontally, but there is acceleration (downward) vertically. A bullet fired is accelerated horizontally by the gunpowder exploding behind it, but once it leaves the barrel then it is just coasting, under the influence of gravity alone (which is downward). While it is coasting, it is considered to be a projectile.
At the start of a problem, you need to choose which direction you will call positive. You can call up positive and down negative, or you can call down positive and up negative. Pick one, and just stick with it through the whole problem. If up is positive, then the acceleration due to gravity is negative. If down is the positive direction, then the acceleration due to gravity is positive.
I understand your reasoning. The problem is how theta is defined. In this context, theta is the angle measured relative to horizontal. The equations Vx = V cos Θ Vy = V sin Θ assume that theta is the angle measured relative to the horizontal. If you called theta the angle relative to vertical, you could still solve the problem, but the sine and cosine would get switched. Hope that helps! Derek Owens
@omar3211 Acceleration near the earth's surface, due to gravity, is 9.8 m/s^2 (or very close to that). That's basically a constant (if you're on earth).
@MsBiebaholic Yes, that is correct. The two equations you mention are actually the same, since horizontally there is no acceleration so the 1/2 a t^2 term reduces to zero. The larger equation simply reduces to the smaller in this case. I don't know an any easy way to memorize the equations of motion, but even if it's just by brute force or practicing, memorizing them is certainly a good idea.
omg you are soooooo helpful!!!!!!! im so glad i found your videos! thanks for taking the time to make these. you dont understand how helpful these are. thank you!
@SohrabR93 It depends on how the problem is set up. Typically a problem can be set up with up being the positive direction (in which case gravity is -9.8), or with down being positive (in which case gravity is +9.8). It can be done either way, as long as you pick one way and stick with it consistently through the whole problem.
@Star123Euro In this problem, the object is a projectile, which means its motion influenced by gravity only. And gravity pulls straight down. The force of gravity does not have any horizontal component, so the horizontal acceleration is zero as long as it is in free flight.
Derek Owens - Thank you so much! Resolving in the Y direction makes so much sense - then you can find the time it would take to drop. This idea links nicely to your last video!!
@juschecknin because gravity doesn't affect it because it's going horizontal, that is why there is no horizontal acceleration, but if it's vertical then yes, because gravity pulls down, gravity doesn't work from side to side (horizontal) therefore there is no acceleration. As far I have seen, every time acceleration is mentioned I think of gravity working on the vertical axis, pulling down, never it acts sideways, unless there is another force of acceleration. I think it goes like that :/
@HairtUB It depends on how the problem is set up. If it is set up consistently, then y and a will both have the same sign in that equation, and there would be no negative square root.
These were difficult at first but after watching these videos I understood the main concepts. I became an expert in projectile motion questions. Thank you so much for doing this :)
@lidyaFACE There are usually two (or more) ways to set up a problem. The acc. can be either positive or negative, depending on how it is set up. For a projectile, though, the horizontal and vertical motions are always independent, and the acceleration of gravity only applies to the vertical motion. For a projectile, the horizontal acceleration will be zero.
im just gonna leave this here... cuz im not too sure if everything required here is in the vid. 1. A ball rolls off a table that is 1.5 m high and lands on the floor, 4.0 m away from the table. a. How long is the ball in the air? b. With what horizontal velocity did the ball roll off the table? c. What is the vertical velocity of the ball just before it hits the floor? d. What is the horizontal velocity of the ball just before it hits the floor?
It's easier to use the equation Vo = range x square root of gravity divided by 2 x the height... i.e. 2 time the height is 200... 9.8 divided by 200 is 0.049... square root of 0.049 is 0.22135. times that by 95 (the range) and you get the answer 21.029. No need to find the time.
Your tutoring skills is very good, but there's still a thing which I would like to clarify: when will you know that acceleration due to gravity is positive or negative? In what circumstances? because every in every Physics class, they will use positive and negative values of acceleration due to gravity. Please help. Thanks in advance.
10x a million ma 2 morro i got a test aout thease and i was sick all week no idea what to do but now i got a clue 10x m* ur da best vry good explenation btw
Great vid as always! I have a question however (hope it's not too foolish): Why do we use the same t in the horizontal equation as the one we used in the vertical one? Maybe I interpret it wrong but doing so wouldn't it mean that it takes the same amount of time to move in a horizontal and vertical way independently? And if so, how do we know it does this? Hopefully I managed to explain my misunderstading well.
@anoorcc Not in this case. If it were a symmetrical parabola, going up and then back down, then we could double the t to find the whole time. In this case, though, we only have half of such a parabola.
Hi Derek. I got a question. An object fall to the earth at 9.8 meter per second. If the vertical distance is 100 meters, then 100 / 9.8 = 10.20 seconds. It would take 10.20 seconds for an object to fall 100 meters. Your calculation showed 4.52 seconds. Can you explain the discrepancy ?
Nice vid, somewhere there is either a mistake in my books or with my interpretation of the formula. If I use the formula "delta x =[(Vf+Vo)/2*delta t]" i get exactly double your answer. That is if I use Vf as 0. Is using Vf as 0 wrong while working with projectiles or is the derivative formula I'm using wrong?
@804YankeeFan The buoyant force is very tiny, and is generally considered small enough to ignore. One could, though, include both the buoyant force and the air resistance in the calculations. The calculations get beyond the scope of this course, though.
EpicJelly Yes, in reality it does lose some horizontal speed due to air resistance. If we neglect the air resistance, though, then the horizontal speed is constant, because gravity only acts in the downward direction, and does not speed the object up or slow it down horizontally. There are no horizontal forces, so there is no change in the horizontal speed. We could include the air resistance, but that changes the problems to a much more difficult problem. In this case we have made the simplifying assumption that air resistance can be neglected.
HI! IN THIS CASE THERE ARE TWO VELOCITIES TO BE CONSIDERED. ONE THAT IS ZERO (THE ONE GOING DOWN). AND THE ONE THAT YOU FOUND OUT. THIS VELOCITY IS """"CONSTANT"""" AND GOING HORIZONTALLY!!!!!!!
My only doubt is that if acceleration horizontally is zero the projectile should keep moving in the same direction (if its velocity is constant ) then why does it stop? ( i m sorry if its 2 obvious)
i think it was a lil bit too complicated. simply just find v initial first which is root 2*g*h. then use then formular v= u + at to get ur time. then substitute for speed = distance/time
Hi hope you won't mind me asking but I am a physics teacher and I'd like to do this kind of thing to support my students remotely during this pandemic.. I am not trying to compete with you but I wondered if you could tell me what hardware / software you use to produce these high standard tutorials. No substitute with physics to being able to write equations draw diagrams etc. Many thanks for any help you can give.
@derekowens So, I have a last question I was wandering about. So are satellites orbiting the sun with the Earth while they are orbiting the Earth? Does this cause the Earth to be an inertial reference frame when calculating its orbit?
You could solve this much faster just by setting the horizontal formula (free-fall) equal to the vertical formula (kinetic) and plugging in your variables.
The setup could be done a few different ways. For example, you could call the height at ground level zero, and the height at the top of the cliff 100, and then up would have to be the positive direction. In this case, I think the problem is slightly easier to set up with down being positive, and the starting height is zero. Hope that helps!
If its motion is influenced by gravity only and gravity pulls straight down, why does the projectile travel 95m? I mean, if you dropped a bullet off the edge of the cliff it would fall straight down, but if it was accelerated out of a gun it would travel several hundred metres due to the acceleration. So in theory this projectile in the video must've had acceleration as opposed to just dropped off the edge?
@ derekowens; When finding the initial horizontal velocity, should time be doubled @ 7:21 so that it account for the whole horizontal time. So instead of 4.52 should it be 9.04. Can anybody else see what I'm talking about as well?
Can't you just use a single equation? because t=x/vcos(θ) y=xtan(θ)-gx^2/2v^2cos^2(θ) tan(0)=0 so: y=gx^2/2v^2cos^2(0) cos^2(0)=1 so: y=gx^2/2v^2 100=9.8((95)^2)/2v^2 If I'm not mistaken. Simply state time in terms of x. "x" velocity never changes in this kind of system after all!
No you treat them as different the only thing common is the time which is to be calculated via downward component and horizontal component remains unchanged so use the calculated time in it I did it the other way used time outta accelerate motion and utilized it is horizontal
dude thank you so much I was having a mental breakdown because of physics but you made it so much more simple
A reminder that 4 years ago you had a mental breakdown over physics. how does it feel now that it is over.
I'm currently going thorough the mental breakdown
@@dahstroyer mental breakdown club
Now I’m having my mental breakdown currently
@@keronrampersad1021 bro same, we have a test tmr. im sobbing i still dont know how to solve it
I never write comments so when I do you know I mean this so sincerely. Physics has never ever made sense to me since middle and high school. Now in college I gotta get it so I don't fail these classes. This is the first time I've felt confident and actually understood even slightly what I was doing. So thank you Derek!!
13
years later, this is still helpful. Thank you sir
Wow the quality and organization of his calculations as well as pace and verbal description of the problem beats even KHANA. Great quality content I'm subbed
love the fact that its colourful!
helps to clearify things a lot!
seriously, i think people like u deserve a great prize... ive physics exam tommorow and i didnt know anything... thanks very much
Absolutely AMAZING! I can't get my Physics teacher to explain any of this to me and my book is too confusing to help me with these kind of problems. Thank you so much for doing this video! I understand it so much more now.
Been 9 years.. how did you do?
Yes, that's correct. That's because for a projectile, there is no acceleration horizontally, so the velocity is the same the entire time, and you can find it with the equation x/t.
Thank you Mr. Owen! I've been frustrated for hours LITERALLY. This is the best video out there! 10/10!
Thank you! I've been having so much problems understanding this and I have my first exam tomorrow.. This helped so much!
Thank you so much Im doing my first year of biomed but i didnt learn much about physics in high school. I know theres projectile formulas but I've been trying to find videos where they use the kinematics one since that increases your understanding. Thank you so much for making these, it was very clear!
I typically use v_0 for the initial velocity. That's v with a zero in the subscript position. It basically means "velocity at time zero" or "velocity when t=0". Then I use v by itself for velocity at some later time. Using u works also, though. Good notation definitely helps, but it's not as important as the concept.
The key here is to distinguish between the vertical and the horizontal. Once the bomb is released, it is a projectile, and there is no horizontal acceleration. There is vertical acceleration, though: gravity is pulling it down.
It also gets more complicated when you include air resistance. Air resistance introduces some horizontal and vertical forces, which depend on the speed.
this video saved my life!! got a test tomorrow morning, but I finally understand now!
Been 9 years how was the test? Did you graduate yet?
I m indian and from half an hour I was dealing with the concept that u made me understand in few mins thanks a lot
@Ell4Sh Yes, that approach would also work. There are typically multiple ways to set up a problem, all of which should give the same final answer. I set it up with 0 a the top because all of the motion and the acceleration are downward, in this problem, and setting it up this way makes all the displacement, velocity, and acceleration numbers positive. The other approach is fine, though.
One of the key points of projectile motion is that the horizontal motion is independent from the vertical motion. There is no acceleration horizontally, but there is acceleration (downward) vertically.
A bullet fired is accelerated horizontally by the gunpowder exploding behind it, but once it leaves the barrel then it is just coasting, under the influence of gravity alone (which is downward). While it is coasting, it is considered to be a projectile.
I'm brushing up on my physics for the MCAT and these videos have been great. Thanks for uploading Derek, I really appreciate it :)
how was mcat!
Been 9 years what did you do? How’s everything?
At the start of a problem, you need to choose which direction you will call positive. You can call up positive and down negative, or you can call down positive and up negative. Pick one, and just stick with it through the whole problem. If up is positive, then the acceleration due to gravity is negative. If down is the positive direction, then the acceleration due to gravity is positive.
I understand your reasoning. The problem is how theta is defined. In this context, theta is the angle measured relative to horizontal.
The equations
Vx = V cos Θ
Vy = V sin Θ
assume that theta is the angle measured relative to the horizontal.
If you called theta the angle relative to vertical, you could still solve the problem, but the sine and cosine would get switched.
Hope that helps!
Derek Owens
@omar3211 Acceleration near the earth's surface, due to gravity, is 9.8 m/s^2 (or very close to that). That's basically a constant (if you're on earth).
I am at the verge of failing Physics class. Chemistry was a mess and I do not want to Jeopardize my GPA! Thank you so much for your help!
Thank you so much for uploading these videos. You explain things so well, and your visuals are amazing.
@MsBiebaholic Yes, that is correct. The two equations you mention are actually the same, since horizontally there is no acceleration so the 1/2 a t^2 term reduces to zero. The larger equation simply reduces to the smaller in this case. I don't know an any easy way to memorize the equations of motion, but even if it's just by brute force or practicing, memorizing them is certainly a good idea.
omg you are soooooo helpful!!!!!!! im so glad i found your videos! thanks for taking the time to make these. you dont understand how helpful these are. thank you!
Thank you so much for making these videos! They made studying for my exam so much easier. :)
@SohrabR93 It depends on how the problem is set up. Typically a problem can be set up with up being the positive direction (in which case gravity is -9.8), or with down being positive (in which case gravity is +9.8). It can be done either way, as long as you pick one way and stick with it consistently through the whole problem.
This literally just saved my whole grade, thank you
Bro your teachings are very smart and easy
Thank you very much
@Star123Euro In this problem, the object is a projectile, which means its motion influenced by gravity only. And gravity pulls straight down. The force of gravity does not have any horizontal component, so the horizontal acceleration is zero as long as it is in free flight.
Derek Owens - Thank you so much!
Resolving in the Y direction makes so much sense - then you can find the time it would take to drop. This idea links nicely to your last video!!
holy shit, i finally UNDERSTAND. Jesus christ. I am jumping with joy right now. Thank you for making sense. Thank you. Seriously.
This is very helpful for my AP class. Thanks so much
@804YankeeFan Yes, that's basically correct. When I throw a ball into the air, the rotation of the earth does not cause the ball to be "left behind".
thanks WOW! your 8 min tutorial rlly helped me do a problem that wasnt even the same as this, so basically you improved my understanding :)
Thanks for posting this. Helped me understand it easier and made my life a lot easier.
@juschecknin
because gravity doesn't affect it because it's going horizontal, that is why there is no horizontal acceleration, but if it's vertical then yes, because gravity pulls down, gravity doesn't work from side to side (horizontal) therefore there is no acceleration. As far I have seen, every time acceleration is mentioned I think of gravity working on the vertical axis, pulling down, never it acts sideways, unless there is another force of acceleration. I think it goes like that :/
@HairtUB It depends on how the problem is set up. If it is set up consistently, then y and a will both have the same sign in that equation, and there would be no negative square root.
These were difficult at first but after watching these videos I understood the main concepts. I became an expert in projectile motion questions. Thank you so much for doing this :)
@lidyaFACE There are usually two (or more) ways to set up a problem. The acc. can be either positive or negative, depending on how it is set up. For a projectile, though, the horizontal and vertical motions are always independent, and the acceleration of gravity only applies to the vertical motion. For a projectile, the horizontal acceleration will be zero.
You sir, are a life saver!
perfect explanation, i am sure i've gained something from it....
im just gonna leave this here... cuz im not too sure if everything required here is in the vid.
1. A ball rolls off a table that is 1.5 m high and lands on the floor, 4.0 m away from the table.
a. How long is the ball in the air?
b. With what horizontal velocity did the ball roll off the table?
c. What is the vertical velocity of the ball just before it hits the floor?
d. What is the horizontal velocity of the ball just before it hits the floor?
Thank you for posting this is helping me study for exams
It's easier to use the equation Vo = range x square root of gravity divided by 2 x the height... i.e. 2 time the height is 200... 9.8 divided by 200 is 0.049... square root of 0.049 is 0.22135. times that by 95 (the range) and you get the answer 21.029. No need to find the time.
@alkhor999 I typically use x to indicate the horizontal position and y to indicate the vertical position.
Your tutoring skills is very good, but there's still a thing which I would like to clarify: when will you know that acceleration due to gravity is positive or negative? In what circumstances?
because every in every Physics class, they will use positive and negative values of acceleration due to gravity. Please help. Thanks in advance.
thanks man,grate explanation,but i don't know why in some cases other lecturers put gravity as -0.98 but it wasn't negative here!
you are amazing!, ths tutorial helped me with a problem that i've been struggling with for days! :)
10x a million ma 2 morro i got a test aout thease and i was sick all week no idea what to do but now i got a clue 10x m* ur da best vry good explenation btw
Great vid as always! I have a question however (hope it's not too foolish): Why do we use the same t in the horizontal equation as the one we used in the vertical one? Maybe I interpret it wrong but doing so wouldn't it mean that it takes the same amount of time to move in a horizontal and vertical way independently? And if so, how do we know it does this? Hopefully I managed to explain my misunderstading well.
@anoorcc Not in this case. If it were a symmetrical parabola, going up and then back down, then we could double the t to find the whole time. In this case, though, we only have half of such a parabola.
Good video. It help me to understand better in this chapter. Good job!
thanks for helping me I was on the verge of crying
T=root 2h/g
U will get t and then
Sx = Uxt
Puting the value of t in above equation u will U
Derek, Nice job explaining this problem!!!! What software/hardware do you use for your chaulboard?
Thanks man have finally made it in physics
your hand writing is beautiful! :)
thank you very much for this great video, it really helped me all .
WOW you are the best!!! Way better the over rated khanacademy
your d bosss...... u deserve a grammy mi bosss
It's very helpful, thankyou for the video
@emmalainesmiles you'd use the equation y = yo + vo(t) + 1/2a(t^2) and plug in what you know for the vertical component to solve for t
Hi Derek. I got a question. An object fall to the earth at 9.8 meter per second. If the vertical distance is 100 meters, then 100 / 9.8 = 10.20 seconds. It would take 10.20 seconds for an object to fall 100 meters. Your calculation showed 4.52 seconds. Can you explain the discrepancy ?
Why is Y positive when it goes downwards? I always see people making Y positive upwards. Now this made me more confused😔
Did you get the reason?
Nice vid, somewhere there is either a mistake in my books or with my interpretation of the formula. If I use the formula
"delta x =[(Vf+Vo)/2*delta t]" i get exactly double your answer. That is if I use Vf as 0. Is using Vf as 0 wrong while working with projectiles or is the derivative formula I'm using wrong?
@804YankeeFan The buoyant force is very tiny, and is generally considered small enough to ignore. One could, though, include both the buoyant force and the air resistance in the calculations. The calculations get beyond the scope of this course, though.
Thank you I couldn't understand until I watch this
Did you pass? It been 9 years
thanks to you, i am going to pass my physics midterm tomorrow
Did you pass? It been 11 years!
thank you very much for this great video, it really helped me allot
Thank you very much. It helps me a lot.
Got it perfectly right! Thank you
thank you so much :) i solved it before you started solving it and my answer was the same as yours :D
@jojosh234
the vertical distance should be at negative because it's going down to Y axis,
how is there no horizontal acceleration? the horizontal acceleration should be negative since it slows down as the object loses it's speed?
EpicJelly Yes, in reality it does lose some horizontal speed due to air resistance. If we neglect the air resistance, though, then the horizontal speed is constant, because gravity only acts in the downward direction, and does not speed the object up or slow it down horizontally. There are no horizontal forces, so there is no change in the horizontal speed. We could include the air resistance, but that changes the problems to a much more difficult problem. In this case we have made the simplifying assumption that air resistance can be neglected.
thanks a lot!
OMG TY FOR ASKING THIS QUESTION. AND THANK YOU FOR THE ANSWER, DEREK.
nice, thanks
Awesome! thanks for the help! please keep on making physics videos, they're a ton of help! 5stars!
@RuthBuzzzzz It would be negative, but when you square anything its always the absolute value of that, so the negative or positive wouldn't matter
I chose up as positive, meaning gravitational acceleration is negative but i get a different answer, -4.52 seconds and -21 m/s ...why is it so?
HI! IN THIS CASE THERE ARE TWO VELOCITIES TO BE CONSIDERED. ONE THAT IS ZERO (THE ONE GOING DOWN). AND THE ONE THAT YOU FOUND OUT. THIS VELOCITY IS """"CONSTANT"""" AND GOING HORIZONTALLY!!!!!!!
thank you so much ! this was so easy to understand and it also makes sense!!!
My only doubt is that if acceleration horizontally is zero the projectile should keep moving in the same direction (if its velocity is constant ) then why does it stop? ( i m sorry if its 2 obvious)
i think it was a lil bit too complicated. simply just find v initial first which is root 2*g*h. then use then formular v= u + at to get ur time. then substitute for speed = distance/time
I have the first problem as my assignment and im confused because the process and the labels used are different TT
vo in the horizontal is 21m/s what about the initial velocity shouldt it be vox=cos(angle)*vo therefore the angle shoul have been given....just saying
Hi hope you won't mind me asking but I am a physics teacher and I'd like to do this kind of thing to support my students remotely during this pandemic.. I am not trying to compete with you but I wondered if you could tell me what hardware / software you use to produce these high standard tutorials. No substitute with physics to being able to write equations draw diagrams etc. Many thanks for any help you can give.
You need a graphics tablet (or Apple Pencil if you have an IPad) and any type of software like OneNote or even Photoshop
I watch these because my teacher is bad
you explained very well. :)
Why the the acceleration is not negative if it's going horizontally?
if the origin is 0 then why horizontal displacement isn't -100m?
@derekowens So, I have a last question I was wandering about. So are satellites orbiting the sun with the Earth while they are orbiting the Earth? Does this cause the Earth to be an inertial reference frame when calculating its orbit?
You could solve this much faster just by setting the horizontal formula (free-fall) equal to the vertical formula (kinetic) and plugging in your variables.
I am quite confused,why is the initial height of the ball vertical zero yet it is thrown from a 100 m heigh building?😢
The setup could be done a few different ways. For example, you could call the height at ground level zero, and the height at the top of the cliff 100, and then up would have to be the positive direction. In this case, I think the problem is slightly easier to set up with down being positive, and the starting height is zero. Hope that helps!
If its motion is influenced by gravity only and gravity pulls straight down, why does the projectile travel 95m?
I mean, if you dropped a bullet off the edge of the cliff it would fall straight down, but if it was accelerated out of a gun it would travel several hundred metres due to the acceleration.
So in theory this projectile in the video must've had acceleration as opposed to just dropped off the edge?
Thank you job well done ✅
i do not understand why the acceleration is positive in this problem! i thought because of gravity pushing down 9.8 would always be negative.
@ derekowens; When finding the initial horizontal velocity, should time be doubled @ 7:21 so that it account for the whole horizontal time. So instead of 4.52 should it be 9.04. Can anybody else see what I'm talking about as well?
Can't you just use a single equation?
because t=x/vcos(θ)
y=xtan(θ)-gx^2/2v^2cos^2(θ)
tan(0)=0 so:
y=gx^2/2v^2cos^2(0)
cos^2(0)=1 so:
y=gx^2/2v^2
100=9.8((95)^2)/2v^2
If I'm not mistaken.
Simply state time in terms of x. "x" velocity never changes in this kind of system after all!
why is the intial velocity horizontally 0? wouldnt it need some velocity once it reached the edge of the cliff to keep moving horizontally.
Isn't the horizontal is the x?
and the vertical is the upward / the y?
I'm confused.
Yup. You are correct.
No you treat them as different the only thing common is the time which is to be calculated via downward component and horizontal component remains unchanged so use the calculated time in it I did it the other way used time outta accelerate motion and utilized it is horizontal
Thanks for posting. But, why did you derive t=squ 2(y)/a??? there was no reason to do that. the original y=1/2at^2 was fine.