It's amazing how this channel covers all concepts you're expected to know by your undergraduate. It covers everything from high school to undergraduate (Almost)! Just Great! 😃
I have a test Friday on one dimensional motion, two dimensional motion and Newton's second law and ive gone through all your sets of videos for those types of problems. Thank you so much. You explain everything very well.
Mr Biezen thank you a lot!!!!!!! you are such a great help. The mark I am scoring now are incredibly increasing And all due to your work and keen on helping us. May God bless you!
Just a question, in my class we're supposed to draw the diagram according and in reference to the x-y plane, I'm I supposed to take the angle (-30°) or (330°) ? or that doesn't matter and we just take the 30° angle?
you can do either -30 or 330 they'll give the same answer, however if you use 30, remember to change your velocity component to a negative as in this example the 30 degrees is negative.
Sir, I have a question. In the last video, even though the ball was still going downward, you took the initial velocity in the y direction as 10m/s. But in this video, you said that since the ball is traveling down wards, its velocity in the y direction should be -10m/s. Why wasnt it also negative 10 in the last video? Both were cases of the ball going downward. Thank you very much. Your videos are phenomenal.
In the previous video, the ball was thrown at an angle upwards above the horizontal, and therefore the initial velocity in the y-direction was positive. Here the ball was thrown at an angle below the horizontal, thus the initial velocity in the y-direction is negative.
We have one video that may help: Differential Equation - 1st Order Solutions (8 of 8) How to Calculate Parachutist's Terminal Speed th-cam.com/video/DS78wFkrBmE/w-d-xo.html
Thank you very much for the videos! And I have a question. Can we multiply g by negative since the ball is moving downward and get a positive g value of 9.8m/s^2 ?
Where did the second equation you use to find the time come from? In my text-book, the closest match is in the form of ; Sy=Uyt + 0.5gt^2. Even if I rearrange to the general form of a quadratic and substitute, -10,-9.8 and 50; I obtain a different equation to yours, resulting in a different answer.
You could use either one, but typically, when we deal with projectiles we compare the direction of the projectile with the axis that indicated the range.
When using the equations of kinematics, we need to be cognizant of the directions. Since g is the acceleration due to gravity and this acceleration is directed downward, g = - 9.8 m/sec^2
Sir, why did we take 'g' negative while calculating vertical motion? Initial velocity direction is already downward. When we throw the stone upward then we take g negative but when it stops going up and starts going down, we take g positive for this free fall as it is in downward direction then shouldn't it be positive here too ( Although 't' is an imaginary number when we keep it positive, so it should be negative but I did not get it concept wise.)
g is ALWAYS - 9.8 m/sec^2 regardless what the ball is doing, moving up, moving down, being motionless at the top, moving sideways, g is always - 9.8 m/sec^2 Remember that F = ma such that the force and the acceleration are always in the same direction. The acceleration due to gravity is always directed downward because the force of gravity is always directed downward.
+Michel van Biezen I really hope you have time, if you were calculating initial velocity in your diagram would v final be 0 since it hit the ground or would it be an unknown velocity before that moment in reallife?
Angles cannot be negative. (for example you will never see a triangle with a negative angle). The negative can be used if you want to reference a direction relative to another direction on the polar coordinate system. (Like 30 degrees below the x-axis can be labeled as -30 degrees relative to the positive x-axis.
pro, is there a special and simplified equation for calculating the distance ( the length of the curve or its trajectory )for the object travelling in a two dimensional where the a is constant without using sqrt((f`(t))*2 + (g`(t))*2 ) dt where x = f(t) , y=g(t) like we use x(t) = v_0t + 1/2at*2 for a is constant in linear motion.
I always recommend that one learns the general approach to a type of problem rather than trying o memorize special equations for specific circumstances or conditions. The methods shown in the video are very standard for this type of problem.
Sir in case a is projectile fired from a height which is unknown but the max range is 'x' and we are asked to find the height for that maximum range how can we find it ?? My text book says that it will be dx/dh=0 but why is it so ??
+Rahul Tiwari You must first be given the angle along which the projectile was fired. Once you know that you solve for the time it takes the projectile to reach the ground and the time for the projectile to reach the maximum range and the you set those 2 times equal to each other to find the height.
Sir its a horizontal projectile motion where I don't have the angle of projectile but I have the initial velocity and using that I got an equation where range is a function of height
Sir I got the exact same thing in the Torricelli's theorem as well where max. range for the function R=√2h(H-h) was calculated by taking dr/dh = 0.But why to take the derivative to be zero while calculating the maxima ??and how can I know whether it is maxima or minima ????
Physics - Mechanics: Motion In Two-Dimension: (12 of 21) Example 1: Plane Dropping Object . the video ive mentioned you took the velocity as it is you didnt use the sin cosine rule why did you use it this time arent they the same question? and iknow in this video that vertically is 20sin30 and horizontally its 20cos30 in this video
I am not sure what you mean by the "sin cosine rule". But rather than use any specific rules, I tend to use the logical concept that you first calculate the time it takes for the object to reach the ground (depends on the y-component of the velocity), and then to use the concept that the time in the y-direction must be equal to the time in the horizontal direction. That works for most projectile problems.
You mentioned in that video I wrote the comment with no vertical initial velocity also no velocity at an angle but you took it this time I hope I could have said it better
I think I understand your question. Note that in this video the initial velocity of the ball was 20 m/sec, and I calculated the initial velocity in the x and y direction (17.3m/sec and 10m/sec), using the sin and cos of the angle.
Could you help me on reso living an exercise like that? I have a mass of 1 kg that receives a force of 6 Nw horizzontally. The mass is on the wall high 20 meters. Where is the distance where the mass goes and the time to arrive in the ground? It,s for my son.....Many thanks. Alberto
We have to make a few assumptions. First we can find the time in the air: y = (1/2) g t^2 therefore t = sqrt (2 * 20 / g) = 2.02 seconds. Then we assume that the force (6 N) is applied horizontally for the entire 2.02 seconds. (is that correct?). Since f = ma a = F/m = 6 m/sec^2 Then the distance x = (1/2) a t^2 = (1/2) (6) (2.02)^2 = 12.2 m.
+jamal jamal When working with the equations of kinematics, up is positive and down is negative. Since the acceleration due to gravity acts in a downward direction g is considered as - 9.8 ms/sec^2.
Thank you sir..but as we know, for falling bodies V = Vo + gt and for bodies thrown up against the gravity V = Vo - gt....falling is negative....but here.... :/
my bad....but if for falling its +9.8... here its also falling downwards.....then why here its -9.8? But thnaks anyways sir.Your tutorials are awesome. :)
t = time and u = initial velocity I meant to say if we use v=u(v initial)+gt here , we would get t=u/g and that works in projectile started and ending at same verticle Height
When used in the equations of kinematics, g is negative, since we essentially work with vectors. (up is positive and down is negative). When the magnitude of g is used then is should be positive. (The magnitude of a vector cannot be negative). (Only the direction of a vector can be negative).
It's amazing how this channel covers all concepts you're expected to know by your undergraduate. It covers everything from high school to undergraduate (Almost)! Just Great! 😃
I have always loved this channel and always will
I have a test Friday on one dimensional motion, two dimensional motion and Newton's second law and ive gone through all your sets of videos for those types of problems. Thank you so much. You explain everything very well.
i don't know why but i love this lecturer. :)
Mr Biezen thank you a lot!!!!!!!
you are such a great help.
The mark I am scoring now are incredibly increasing
And all due to your work and keen on helping us. May God bless you!
Just a question, in my class we're supposed to draw the diagram according and in reference to the x-y plane, I'm I supposed to take the angle (-30°) or (330°) ? or that doesn't matter and we just take the 30° angle?
you can do either -30 or 330 they'll give the same answer, however if you use 30, remember to change your velocity component to a negative as in this example the 30 degrees is negative.
You are an amazing tutor i understand projectile way better now
I'm not yet sure but I think I started to like physics! thank youuuuuu so much!
thankyou so much for these videos, so easy to understand and great examples.
Sir, I have a question. In the last video, even though the ball was still going downward, you took the initial velocity in the y direction as 10m/s. But in this video, you said that since the ball is traveling down wards, its velocity in the y direction should be -10m/s. Why wasnt it also negative 10 in the last video? Both were cases of the ball going downward.
Thank you very much. Your videos are phenomenal.
In the previous video, the ball was thrown at an angle upwards above the horizontal, and therefore the initial velocity in the y-direction was positive. Here the ball was thrown at an angle below the horizontal, thus the initial velocity in the y-direction is negative.
this is very helpfull thank man you know i am watching this 45 minute before my mid term test
appreciate this channel. god bless you!
Thanks Michel. Do you have a hint were to find the same problem but factoring drag linearly proportional to the projectile's velocity? -- Thanks.
We have one video that may help: Differential Equation - 1st Order Solutions (8 of 8) How to Calculate Parachutist's Terminal Speed th-cam.com/video/DS78wFkrBmE/w-d-xo.html
thank you! this helped me a lot. it didnt included in my freaking textbook n i hv upcoming final exam
I appreciate the help. I needed it. Badly.
Thank you very much for the videos! And I have a question. Can we multiply g by negative since the ball is moving downward and get a positive g value of 9.8m/s^2 ?
1:55 Why G is taken negative? Doesn't V(y) direction is towards the gravity?
When using the equations of kinematics, g is always - 9.8 m/sec^2
Michel van Biezen- Aah..I messed up the sign convention. Thank you.
Why not just use the angle in the 4th quadrant : 330 degrees to make the y component of velocity negative from the start
In physics we typically find an angle between 0 and 90 degrees relative to the vertical or horizontal.
I'm taking PHYS-211 at RIT this coming up fall. Could you please come teach here? :D
Where did the second equation you use to find the time come from? In my text-book, the closest match is in the form of ; Sy=Uyt + 0.5gt^2.
Even if I rearrange to the general form of a quadratic and substitute, -10,-9.8 and 50; I obtain a different equation to yours, resulting in a different answer.
he used the quadratic equation to solve after he put in values for kinematic equation
How would I solve if I was given the horizontal displacement but not initial velocity?
The time to reach the ground is equal to the time to reach the distance. (Set the 2 equations equal to each other).
Thank you. Your videos are really really helpful..Good Job!
Sir , I am wonder that in the previous video the velocity in vertical direction is positive but in this video is negative
+Abdulrhman ALfahim
The initial velocity (in the y-direction) is negative in this problem.
thanks sir I understand it
Sth i am curious about why do we always use the angle between V and the x axis not the y axis?
You could use either one, but typically, when we deal with projectiles we compare the direction of the projectile with the axis that indicated the range.
why is the angle for calculating the vertical velocity 30, not 60. because if you think about it, 90-30 is 60
The angle is relative to the horizontal.
Thank you for your help
Happy to help
hi..isn't [g] suppose to be positive? because gravity is downwards
When using the equations of kinematics, we need to be cognizant of the directions. Since g is the acceleration due to gravity and this acceleration is directed downward, g = - 9.8 m/sec^2
why in this case a = -9.8, so how about the acceleration value when we throw a ball up straight
The acceleration in the vertical direction will always be: a = g = - 9.8 m/sec^2
gravity is always taken as -9.8 as gravity acts in the downward direction and we usually take the upward direction as positive.
Sir, why did we take 'g' negative while calculating vertical motion? Initial velocity direction is already downward.
When we throw the stone upward then we take g negative but when it stops going up and starts going down, we take g positive for this free fall as it is in downward direction then shouldn't it be positive here too ( Although 't' is an imaginary number when we keep it positive, so it should be negative but I did not get it concept wise.)
g is ALWAYS - 9.8 m/sec^2 regardless what the ball is doing, moving up, moving down, being motionless at the top, moving sideways, g is always - 9.8 m/sec^2 Remember that F = ma such that the force and the acceleration are always in the same direction. The acceleration due to gravity is always directed downward because the force of gravity is always directed downward.
@@MichelvanBiezen Thank you sir for clarification.
I love all your videos. But this particular question wrong. Due to you a,b&c signs coefficients
+Carlos Acosta
The video is correct.
wow, really ugh ok... never the less i am a huge fan. i have recommended your channel. there is a lot of views and likes coming through!!
+Michel van Biezen I really hope you have time, if you were calculating initial velocity in your diagram would v final be 0 since it hit the ground or would it be an unknown velocity before that moment in reallife?
sir why do u put the value of time in the x dirrection equation s= vt
why dont we put it in the s =ut +1/2at^2 equation
+Monesa Murryem
Since there is no acceleration in the x-direction, a = 0
why didnt u put the 30° to be -30° since it is below x axis?
Angles cannot be negative. (for example you will never see a triangle with a negative angle). The negative can be used if you want to reference a direction relative to another direction on the polar coordinate system. (Like 30 degrees below the x-axis can be labeled as -30 degrees relative to the positive x-axis.
pro, is there a special and simplified equation for calculating the distance ( the length of the curve or its trajectory )for the object travelling in a two dimensional where the a is constant without using sqrt((f`(t))*2 + (g`(t))*2 ) dt where x = f(t) , y=g(t) like we use x(t) = v_0t + 1/2at*2 for a is constant in linear motion.
I always recommend that one learns the general approach to a type of problem rather than trying o memorize special equations for specific circumstances or conditions. The methods shown in the video are very standard for this type of problem.
Nice channel❤❤
Thank you. Glad you found it.
Thanks. I'm from morroco I like the physics
@@MichelvanBiezenHello, can I come to you to study with you?
I think the best we can do here is to use our videos.
When calculating for initial y velocity or x velocity, how do I state the direction?
Use the sign convention. Up and to the right is positive. Down and to the left is negative.
Sir in case a is projectile fired from a height which is unknown but the max range is 'x' and we are asked to find the height for that maximum range how can we find it ?? My text book says that it will be dx/dh=0 but why is it so ??
+Rahul Tiwari
You must first be given the angle along which the projectile was fired. Once you know that you solve for the time it takes the projectile to reach the ground and the time for the projectile to reach the maximum range and the you set those 2 times equal to each other to find the height.
Sir its a horizontal projectile motion where I don't have the angle of projectile but I have the initial velocity and using that I got an equation where range is a function of height
Sir I got the exact same thing in the Torricelli's theorem as well where max. range for the function R=√2h(H-h) was calculated by taking dr/dh = 0.But why to take the derivative to be zero while calculating the maxima ??and how can I know whether it is maxima or minima ????
Physics - Mechanics: Motion In Two-Dimension: (12 of 21) Example 1: Plane Dropping Object .
the video ive mentioned you took the velocity as it is you didnt use the sin cosine rule why did you use it this time arent they the same question? and iknow in this video that vertically is 20sin30 and horizontally its 20cos30 in this video
I am not sure what you mean by the "sin cosine rule". But rather than use any specific rules, I tend to use the logical concept that you first calculate the time it takes for the object to reach the ground (depends on the y-component of the velocity), and then to use the concept that the time in the y-direction must be equal to the time in the horizontal direction. That works for most projectile problems.
You mentioned in that video I wrote the comment with no vertical initial velocity also no velocity at an angle but you took it this time I hope I could have said it better
I think I understand your question. Note that in this video the initial velocity of the ball was 20 m/sec, and I calculated the initial velocity in the x and y direction (17.3m/sec and 10m/sec), using the sin and cos of the angle.
Oh thank you Mr. Michel van Biezen
if I don't have the initial v but i have the time (in sec)??
Thank u sir from India
Welcome to the channel!
@@MichelvanBiezen it's my pleasure sir.. your lectures are very helpful. thanks
why are we taking gravity as minus? it should be pozitif because the ball is going to below. it is getting faster, isnt? it?
When using the equations of kinematics, the convention requires down to be negative. Since g is directed downward, we use - 9.8 m/sec^2
@@MichelvanBiezen thanks, you are a real hero :)
Thank you sir very clear explanation.
THANK YOU SIR FOR THE VERY HELPFUL VIDEOS I ALSO LIKE YOUR BOW TIES SIR LOL :)
Could you help me on reso living an exercise like that? I have a mass of 1 kg that receives a force of 6 Nw horizzontally. The mass is on the wall high 20 meters. Where is the distance where the mass goes and the time to arrive in the ground? It,s for my son.....Many thanks. Alberto
We have to make a few assumptions. First we can find the time in the air: y = (1/2) g t^2 therefore t = sqrt (2 * 20 / g) = 2.02 seconds. Then we assume that the force (6 N) is applied horizontally for the entire 2.02 seconds. (is that correct?). Since f = ma a = F/m = 6 m/sec^2 Then the distance x = (1/2) a t^2 = (1/2) (6) (2.02)^2 = 12.2 m.
Many thanks for helping me
should the displacement be negative not positive as it goes down the buliding???
+Masi Alshehab
The magnitude of a vector cannot be negative. A vector can only have a negative direction.
why the gravity is negative?
+jamal jamal
When working with the equations of kinematics, up is positive and down is negative. Since the acceleration due to gravity acts in a downward direction g is considered as - 9.8 ms/sec^2.
+Michel van Biezen Thank you.
why is g negative here?isnt it supposed to be positive when falling or throwing from above to the ground?
g is the acceleration due to gravity and it is always directed towards the ground. Thus it must be negative.
Thank you sir..but as we know, for falling bodies V = Vo + gt and for bodies thrown up against the gravity V = Vo - gt....falling is negative....but here.... :/
No, the equation is always V = Vo + gt (where g = -9.8 m/sec^2) for all situations.
my bad....but if for falling its +9.8... here its also falling downwards.....then why here its -9.8? But thnaks anyways sir.Your tutorials are awesome. :)
How would you solve the same problem if the angle was unknown?
We have examples in the playlists where the angle is not known. PHYSICS 3.5 FINDING ANGLE IN PROJECTILE MOTION
Hello, how did you know to use 30 degrees and not 60? thank you.
The angle was given in the problem.
why is y final zero?
At the end, the object will be on the ground (y = 0)
I think it must be -50
Fire!
why doesn't T= u/g work here ?
What does "T" and "u" represent?
t = time and u = initial velocity
I meant to say if we use v=u(v initial)+gt here , we would get t=u/g and that works in projectile started and ending at same verticle Height
❤❤❤❤
Thank you. Glad you enjoyed it. 🙂
When should I take g as positive and when as negetive
When used in the equations of kinematics, g is negative, since we essentially work with vectors. (up is positive and down is negative). When the magnitude of g is used then is should be positive. (The magnitude of a vector cannot be negative). (Only the direction of a vector can be negative).
thank u sir
I can't get the 32.86, how did you?
Square root of 1080.
😊😊😊
Thank you
Thanks man probs first guy in 2019 lol
thanks
دەستت خۆش بێت
I don't understand any thingv
I love you Russian man
Oh! so funny lecture thanks
😂 😊 😅
Thank you
Прикольно !
Thanks