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
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!
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.
@derekowens I was wandering If I drop a ball in the air, the air's bouyancy force is acting on me and accelerating me with the weight. Am I right? So why dont people consider the extra bouyancy force?
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!
@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?
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.
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.
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.
@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).
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)
Hello, I would like to ask something, say that the projectile is a bomb released from a fighter plane going at a certain speed, is the acceleration of the object still zero, or is it the acceleration?
@ 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?
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?
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!!
@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.
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.
@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.
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.
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?
im really bad at physics im watching this to get a better understanding obviously. i would make the top of the cliff 100m and the bottom 0m.. id also make my a -9.8 m/s.. i guess what im asking is how do you lay it out that way?
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.
@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.
@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.
That's because a projectile (by the definition of projectile) is just coasting through the air. There is no engine, no propeller, so no thrust. The only force on it is gravity. And gravity only acts vertically, downward, so there is downward acceleration, but none horizontally.
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 ?
@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 :/
respected sir you said that vertical initial velocity will be 0 but if we apply vectors initial velocity should be u sin(theta) and as it is a horizontal motion, so theta should be 90. now putting value of theta in u sin(theta) we get u sin 90 which as we know is equal to 1 as sin90=1. So please help me clear my doubt.
@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.
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?
Margaret Ayambem It is just third kinematics equation: Δy=v0t+1/2at^2 We know that: Δy= y - y0 So, y - y0 = v0t+1/2at^2 y = y0+v0t+1/2at^2 www.khanacademy.org/science/physics/one-dimensional-motion/kinematic-formulas/a/what-are-the-kinematic-formulas
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
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!
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
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
13
years later, this is still helpful. Thank you sir
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
Thank you Mr. Owen! I've been frustrated for hours LITERALLY. This is the best video out there! 10/10!
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
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!
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 for uploading these videos. You explain things so well, and your visuals are amazing.
Thank you so much for making these videos! They made studying for my exam so much easier. :)
This literally just saved my whole grade, thank you
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?
how could you determine the way of which should you firstly use horizontal or vertical ?
is it possible to find the speed if the vertical velocity wasn't 0 given the x and y distances only ?
@derekowens why when we take the horizontal the acceleration is zero? please tell me
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 :)
@derekowens I was wandering If I drop a ball in the air, the air's bouyancy force is acting on me and accelerating me with the weight. Am I right? So why dont people consider the extra bouyancy force?
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
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!
Awesome! thanks for the help! please keep on making physics videos, they're a ton of help! 5stars!
@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?
Thanks for posting this. Helped me understand it easier and made my life a lot easier.
you are amazing!, ths tutorial helped me with a problem that i've been struggling with for days! :)
Does the projectile objects have the inertia of the rotation speed of the earth?
why is the intial velocity horizontally 0? wouldnt it need some velocity once it reached the edge of the cliff to keep moving horizontally.
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?
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.
Why the the acceleration is not negative if it's going horizontally?
This is very helpful for my AP class. Thanks so much
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?
What is the software and device do you use to draw this pictures?
for the horizontal can u do velocity=distance/time does it still work?
do you use the equation d=vit+1/2at^2
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.
Derek, Nice job explaining this problem!!!! What software/hardware do you use for your chaulboard?
idk if this is a dumb question but can someone anwser??. How do you know when to rearrange the equation???
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.
why we use -g in projectile motion?
@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).
if the origin is 0 then why horizontal displacement isn't -100m?
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)
Hello, I would like to ask something, say that the projectile is a bomb released from a fighter plane going at a certain speed, is the acceleration of the object still zero, or is it the acceleration?
You sir, are a life saver!
what s the software you use to draw...?
I have the first problem as my assignment and im confused because the process and the labels used are different TT
Bro your teachings are very smart and easy
Thank you very much
@ 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?
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?
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!!
What about when horizontal distance is not given ??
@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.
thank you very much for this great video, it really helped me all .
Good video. It help me to understand better in this chapter. Good job!
It's very helpful, thankyou for the video
thank you very much for this great video, it really helped me allot
holy shit, i finally UNDERSTAND. Jesus christ. I am jumping with joy right now. Thank you for making sense. Thank you. Seriously.
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!
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.
Thanks man have finally made it in physics
Got it perfectly right! Thank you
@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.
Why did you not just use v^2-2aS=u^2 ?
to find the horizontal
your hand writing is beautiful! :)
@alkhor999 I typically use x to indicate the horizontal position and y to indicate the vertical position.
perfect explanation, i am sure i've gained something from it....
Mr Owen, how do I contact you. My daughter wants to take some classes with you. Please let me know. Thanks.
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.
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
thank you so much ! this was so easy to understand and it also makes sense!!!
im really bad at physics im watching this to get a better understanding obviously. i would make the top of the cliff 100m and the bottom 0m.. id also make my a -9.8 m/s.. i guess what im asking is how do you lay it out that way?
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.
Why does your gravity at positive? it should be at negative because it's going to negative Y axis
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?
@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".
Thank you very much. It helps me a lot.
@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.
@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 explained very well. :)
Thank you very much. It helps me a lot
just wanna ask, why is there no accelaration in horizontal motion?
That's because a projectile (by the definition of projectile) is just coasting through the air. There is no engine, no propeller, so no thrust. The only force on it is gravity. And gravity only acts vertically, downward, so there is downward acceleration, but none horizontally.
what is Y?
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 ?
@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 :/
@jojosh234
the vertical distance should be at negative because it's going down to Y axis,
respected sir
you said that vertical initial velocity will be 0 but if we apply vectors initial velocity should be u sin(theta) and as it is a horizontal motion, so theta should be 90.
now putting value of theta in u sin(theta) we get u sin 90 which as we know is equal to 1 as sin90=1. So please help me clear my doubt.
@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.
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?
Is there a video that shows how you derived the formula at the beginning: y= y+vt+1/2at
Margaret Ayambem
It is just third kinematics equation:
Δy=v0t+1/2at^2
We know that:
Δy= y - y0
So, y - y0 = v0t+1/2at^2
y = y0+v0t+1/2at^2
www.khanacademy.org/science/physics/one-dimensional-motion/kinematic-formulas/a/what-are-the-kinematic-formulas
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
thanks for helping me I was on the verge of crying
thank you so much :) i solved it before you started solving it and my answer was the same as yours :D
THanks man!!!! It helps a lot dude
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!
How exactly did you get the a=9.8 m/s²
That's the acceleration due to gravity near earth's surface. That's basically a constant in these problems.
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
Thank you job well done ✅
your d bosss...... u deserve a grammy mi bosss