Wow.. Professor Sama, you really are a ninja of physics.. thank you from Italy.. I am a 40(more than 40) years old guy Who has signed to university and I was looking for a clear explanation of the projectile motion.. thank you very much..
Thank You so much for making this video. It really helped me in my online class, especially self study is not easy. Thank you for sharing your ideas that is very clear and so understandable. God bless and more power!
Thanks alot sir, excellent way of explaining complicated stuff... I had huge doubts in projectile motion, i looked at many videos but only found this one useful and best...
Thank you so much! This video really helped me see the breakdown of the equations used to find the specifics (time, max height, vertical height, etc.) Please keep making videos and thank you :)
The video is really helpful, thank you. But I feel like it would have been more clear if you used numbers on your examples, and can the final range formula be further simplified into (v initial^2*(sin teta*)2)/g because I assume the angles will be the same?
If you are interested in an in-depth analysis of the geometry and the physics of projectile motion I recommend: th-cam.com/video/HPehCUv6bEY/w-d-xo.html&ab_channel=Math%2CPhysics%2CEngineering
Really wish I could hear this video. Turned my TV volume all the way up on the chrome cast, tried AirPods, the sound is so low. I do appreciate the making of it tho! Just a remark
how would i do a question lik this A projectile is fired with an initial velocity of 120 ms-1. The projectile has a time of flight of 18.75s. Determine: a) The range of the projectile; b) The maximum height attained, and the time at which this height is attained;
what if the projectile had acceleration, per say, a rocket. the rocket acceleration is in a changing direction. and the parabola is less circular (longer on the +y path and shorter on the -y). What would be added?
this equation is true for a cannon or snipers sake, but a rocket that dosent have one blast that sets it in motion (rather a continuous acceleration in its trajectory ) is written differently, How?
first we know Y=Yo+Voy-1/2 g t^2.so if t total equal to t flight it means y=h at max point.then h=Vosinθ x Vosinθ/2-(1/2 x g x Vo^2 sin^2θ/g^2) from this equalition (Vo^2 x sin^2θ)/2g=h
Which time(s) and distance(s) are you given? If you have the ∆x for t(max), or position & time of impact, then you should have your Vx for when Vy = 0 because Vx is constant. If your starting height and impact height are the same, the parabola from start to finish will be symmetrical, therefore, its midway point will be half of t(max). Long story short, you have to work backwards, and your max height is tangeant to the highest point on the parabola.
If at a time T the direction of the velocity is at 90 degrees to the initial direction of the velocity is given in a problem what can you derive from that?
my physics exam is tomorrow and i still dont understand projectiles which will most definitely be there but theres so many other units to study theres no time 😭
I need help in calculating the trajectory length. Yes, it's the length of the parabola or projectile not the horziontal or vertical displacement. Any help?
You first need to write y vs x. You can do this by eliminating time. Then you will need to find the arc length. This is a standard math problem. I suggest googling it. It will involve integrating a square root function of dy/dx
In real life world, would it be fair to assume that there would be a deceleration of the motion in x-axis due to air friction? or is it negligible? If negligible, does that mean if i plot the graph of horizontal distance covered over time it will be a straight line and straight to zero when the final Y is 0?
In the real world, you would need to include air resistance as it has a non-negligible deceleration on the x and y motion of the projectile. I can’t remember the formula for drag off the top of my head but it is necessary for the real world.
In linear motion velocity and acceleration are in the same direction but in the projectile motion velocity is upward and acceleration is downward how is that possible could you explain it to me.
Great vid but i have a small question, in my school book it says that time to reach maximum height= -vy divided by g, which is -9.8m/s2, is there a difference between this and the one in the video?
at 14:26 i write an equation for t_top=v_0*sin(theta)/g. This is the same as the equation you wrote. the vy in your equation is the initial y component of the velocity.
@@PhysicsNinja my problem is like this on the video "a ball has been thrown with initial velocity of 28m/s at 30° angle" my delta y displacement is negative did I answer it right? thank you sir
There are indeed two points where ∆y = 10. In this example, one answer will have a positive Vy and the other will have a negative Vy, assuming it isn't the top of the arc.
is the initial height same as the maximum height? cuz i have only have the velocity and the angle, ive already solved for the max height but the time and range i cant seem to solve it because i need the int. height
If you are interested in an in-depth analysis of the geometry and the physics of projectile motion I recommend: th-cam.com/video/HPehCUv6bEY/w-d-xo.html&ab_channel=Math%2CPhysics%2CEngineering
If only our professors/teachers could be as crystal clear as you! Thank God for your brilliant mind. Keep up the excellent work.
Thank you so much
Well explained thanks for your time
This is in fact the single most helpful video on the internet. Thank you.
Glad you think so!
Wow, you saved my life. I’m forever indebted to you.
OA
@@aldrinjames2565Fsfs
Wow.. Professor Sama, you really are a ninja of physics.. thank you from Italy.. I am a 40(more than 40) years old guy Who has signed to university and I was looking for a clear explanation of the projectile motion.. thank you very much..
Thank you so much
This is the best projectile motion video of all time
literally such a simplified and amazingly detailed explanation. I'm absolutely really grateful for it 7 years later🥰💗
Glad it was helpful!
Great Job, thank you. Trying to explain it to my son and came across this video. It reminds me of my teacher
Thank You so much for making this video. It really helped me in my online class, especially self study is not easy. Thank you for sharing your ideas that is very clear and so understandable. God bless and more power!
Thanks alot sir, excellent way of explaining complicated stuff...
I had huge doubts in projectile motion, i looked at many videos but only found this one useful and best...
Thank you so much!! Good luck with your studies. I wish you much success.
@@PhysicsNinja thanks sir...
First video about this subject that I actually understood, and, more importantly, learned something new!
Every time I thought "why is he doing that??" you explained it perfectly.
Such a lifesaver! Better than my professors
Thank you so much! This video really helped me see the breakdown of the equations used to find the specifics (time, max height, vertical height, etc.) Please keep making videos and thank you :)
Watching from Africa and this has made me think smart 😊💯💯💯
I hope you succeed in your studies
If i still dont get it im quiting school
Soo what happened?
probably because he didn't provide an example
Same
Updates?
Best idea bro !!!!!!!!
This 20 mins video is clearer than our 1hr discussion-
You are the reason I did not get an F in my midterm, thank you Physics Ninja
Wooow , tomorrow is my exam and you really saved my life , appreciate it ❤️❤️❤️❤️
Best of luck!
This video made the concept so much easier, i get it now
The video is really helpful, thank you. But I feel like it would have been more clear if you used numbers on your examples, and can the final range formula be further simplified into (v initial^2*(sin teta*)2)/g because I assume the angles will be the same?
If you are interested in an in-depth analysis of the geometry and the physics of projectile motion I recommend:
th-cam.com/video/HPehCUv6bEY/w-d-xo.html&ab_channel=Math%2CPhysics%2CEngineering
Now everything projectile seems like child's play. Dam easy.
Thanks a million 🙏
Thanks for making it clearer. I was literally crying genuine tears because of how cooked i am.
You got this!
I understand everything you said because you know how to reach us😊😊👍👍
Wow u are an amazing professor!! Easily understood
Thank you!
Great vid, in the last equation you can simplify 2 sin theta cos theta to -> sin2theta
This helped me on my exam
~thankyou from India
Really wish I could hear this video. Turned my TV volume all the way up on the chrome cast, tried AirPods, the sound is so low. I do appreciate the making of it tho! Just a remark
True!
5:41 one thing even chat gpt fails to answer me is WHY is there a 1/2 there where the F did that come from?
absolutely phenomenal teaching! Wow
Thank you so much! the explanation is very nice and smooth thank you again sir have a great day ahead
Thank you! you make my physics more easier.
THANK YOU!!!!! it's so clear now, i was so confused. earned a sub
Thank you now I understand this topic wayyy better
May i ask how did you simplify that ymax formula to its simplest form from that more complex ymax formula, i am just confused
I have the max height and range and need to find initial velocity, I’ve been stuck for ages send help
I will post a video tonight on a basketball problem that deals with this. Stay tuned!!
Physics Ninja thank you so much!
how would i do a question lik this
A projectile is fired with an initial velocity of 120 ms-1. The projectile has a time of flight of 18.75s.
Determine:
a) The range of the projectile;
b) The maximum height attained, and the time at which this height is attained;
If the angle is not given, how would you go about solving the problem?
thank you for your effort, patient and very precise explanation.
At 6:28 why above equation before g has 1/2 but below equation doesnt have?
I wish you were my teacher. Thank you so so much.
Wow, thank you!
bro thank god for this man
What if the object is initially displaced upwards?
Thank you so much sir i was studying for the MCAT and i was struggling with this concept but now i understood it properly and much more confident
Good luck with the MCAT
Can you make a video on the equation of trajectory of the projectile motion😊
Yes, maybe this weekend
@@PhysicsNinja thank you sir 🤗.
what if the projectile had acceleration, per say, a rocket. the rocket acceleration is in a changing direction. and the parabola is less circular (longer on the +y path and shorter on the -y). What would be added?
this equation is true for a cannon or snipers sake, but a rocket that dosent have one blast that sets it in motion (rather a continuous acceleration in its trajectory ) is written differently, How?
Plus can I have a detailed explaination on the mathematics involved in deriving max height
first we know Y=Yo+Voy-1/2 g t^2.so if t total equal to t flight it means y=h at max point.then h=Vosinθ x Vosinθ/2-(1/2 x g x Vo^2 sin^2θ/g^2) from this equalition (Vo^2 x sin^2θ)/2g=h
14:40 Ehh, why is V0sin theta squared? As you can see, it is not squared on the velocity formula
Can anyone help pls ?
@@Kvarggistapäivää substitue t top into equation (2) , u'll get Vosintheta x Vosintheta/g ...
Nicely done my brother.
Thank you kindly
nice work but what about the angle at which the particle hits the ground?
Wow this is great
you are simply excellent, this video literally cleared everything...
But what if there's no given angle? The only given are distance and time. How to find the maximum height?
Which time(s) and distance(s) are you given? If you have the ∆x for t(max), or position & time of impact, then you should have your Vx for when Vy = 0 because Vx is constant.
If your starting height and impact height are the same, the parabola from start to finish will be symmetrical, therefore, its midway point will be half of t(max). Long story short, you have to work backwards, and your max height is tangeant to the highest point on the parabola.
@@duttroach8489 what happens if you have no given angle but only have speed and distance/range how would you solve for Vy?
Very clear explanation
Thank you 😊
Hi can I just ask how to solve time in projectile motion? The t=1s t=2s t=3s
which quadratic equation satisfy this path ???
for time up, the formula is velocity x divide by -9.8) , so the time will become -ve value. is it true?
when its physics exam tomorrow and ur so cooked u play the video with 2x speed
I was told that range was |v|^2 * sin(2theta) / g, is this the same as 2Vo^2 * sin(theta) * cos(theta) / g?
Same, using a trigonometric metric identity 2sinxcosx=sin2x
Thank you sensei, how may I repay you for your help.
You really make me proud sir I now know what I did not do
You got this!
If at a time T the direction of the velocity is at 90 degrees to the initial direction of the velocity is given in a problem what can you derive from that?
Sir can you help me. I nedd a some biomechanics problems.( segmemt lenth segment mass , COM )
my physics exam is tomorrow and i still dont understand projectiles which will most definitely be there but theres so many other units to study theres no time 😭
Thank you so much. this is very helpful.
What about the X value when Y is at its max
1/2 of range, since it is symmetrical
How come i now understand ballistic calculations but i still cant understand polar patterns being taught by my precal II teacher
This helped, thank you.
omg so clear tysm 😊
Very helpful!
im in lecture, wit this playing instead , im cooked
I need help in calculating the trajectory length. Yes, it's the length of the parabola or projectile not the horziontal or vertical displacement. Any help?
You first need to write y vs x. You can do this by eliminating time. Then you will need to find the arc length. This is a standard math problem. I suggest googling it. It will involve integrating a square root function of dy/dx
In real life world, would it be fair to assume that there would be a deceleration of the motion in x-axis due to air friction? or is it negligible?
If negligible, does that mean if i plot the graph of horizontal distance covered over time it will be a straight line and straight to zero when the final Y is 0?
In the real world, you would need to include air resistance as it has a non-negligible deceleration on the x and y motion of the projectile. I can’t remember the formula for drag off the top of my head but it is necessary for the real world.
this video was insanely helpful, thank you so much
why is a = -g when g is directed downwards
Down is taken to be the negative direction.
If a body is projected upward then it's acceleration is -ve why??
Gravity is down and we take this direction to be negative
Thanks for answering my question ❤
In linear motion velocity and acceleration are in the same direction but in the projectile motion velocity is upward and acceleration is downward how is that possible could you explain it to me.
you are damn good professor.
You're an iron man, wow, that's great. Huge accomplishment. That jacket is cool.
There’s no greater accomplishment in life than finishing an Ironman.
@@PhysicsNinja you're damn right, sir
CALCULATE THE VELOCITY WITH WHICH THE BALL LEAVESTHE PLAYER HAND IN 5 SECOND? WHEREBY NO DISPLACEMENT. HELP PLEASE
Great vid but i have a small question, in my school book it says that time to reach maximum height= -vy divided by g, which is -9.8m/s2, is there a difference between this and the one in the video?
at 14:26 i write an equation for t_top=v_0*sin(theta)/g. This is the same as the equation you wrote. the vy in your equation is the initial y component of the velocity.
sir I think v is not initial Velocity v is final Velocity and u is intial velocity...
can total delta y displacement can be negative?
Yes, if you drop inside a hole. Negative displacement simple tell you the direction. Negative would down relative to where you started.
@@PhysicsNinja my problem is like this on the video "a ball has been thrown with initial velocity of 28m/s at 30° angle" my delta y displacement is negative did I answer it right? thank you sir
2 sin a cos a = sin 2a. Therefore, max range is when a = 45 degrees.
Thank you teacher❤
G is pointing down so it negative but where’s the 1/2 gt square coming from
the kinematic equation
I have the average speed and the time the projectile takes to reach the ground, I need to calculate the range and maximum height, please help
what if I have to find the time when the y displacement is 10 ?? considering there will be two answers
There are indeed two points where ∆y = 10. In this example, one answer will have a positive Vy and the other will have a negative Vy, assuming it isn't the top of the arc.
What would happen if the initial velocity upon launch is 0? Or is that not possible?
That’s the easiest case- Maximum height=0, Range=0
Thank you! i get it now
is the initial height same as the maximum height? cuz i have only have the velocity and the angle, ive already solved for the max height but the time and range i cant seem to solve it because i need the int. height
is dy the maximum height?
If you are interested in an in-depth analysis of the geometry and the physics of projectile motion I recommend:
th-cam.com/video/HPehCUv6bEY/w-d-xo.html&ab_channel=Math%2CPhysics%2CEngineering
good teaching thnak you
So nice of you
Thanks a lot, sir.
This is 10/10
شكرا جزيلا لك شرح ممتاز جدا
Very interesting 🎉
thank you so much!!!
You're welcome!
Awesome video
Thanks im using this for a program on a rocket
THANK YOU
Thank you 😊