Projectile Motion: Finding the Maximum Height and the Range

[@torilecky488]

This is in fact the single most helpful video on the internet. Thank you.

[@PhysicsNinja]

Glad you think so!

[@metiburussie8973]

If only our professors/teachers could be as crystal clear as you! Thank God for your brilliant mind. Keep up the excellent work.

[@PhysicsNinja]

Thank you so much

[@veronicanjovu8353]

Well explained thanks for your time

[@iRobotray]

This is the best projectile motion video of all time

[@lucasanfilippo5370]

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..

[@PhysicsNinja]

Thank you so much

[@lacertus2753]

Wow, you saved my life. I’m forever indebted to you.

[@aldrinjames2565]

OA

[@CallmeMizami]

​@@aldrinjames2565Fsfs

[@asteriabosco7272]

literally such a simplified and amazingly detailed explanation. I'm absolutely really grateful for it 7 years later🥰💗

[@PhysicsNinja]

Glad it was helpful!

[@lukaround]

Great Job, thank you. Trying to explain it to my son and came across this video. It reminds me of my teacher

[@jessamaecabizares6598]

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!

[@johnmwebela2226]

Watching from Africa and this has made me think smart 😊💯💯💯

[@PhysicsNinja]

I hope you succeed in your studies

[@xxsweatxx6945]

If i still dont get it im quiting school

[@Aetherakabane]

Soo what happened?

[@sseolhi]

probably because he didn't provide an example

[@YutakaQ]

Same

[@Al-Suzuki]

Updates?

[@ZulfAMysam]

Best idea bro !!!!!!!!

[@lilac4196]

Such a lifesaver! Better than my professors

[@HAL_HLA]

This 20 mins video is clearer than our 1hr discussion-

[@burgamin5467]

First video about this subject that I actually understood, and, more importantly, learned something new!

[@ayushagrawal4275]

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...

[@PhysicsNinja]

Thank you so much!! Good luck with your studies. I wish you much success.

[@ayushagrawal4275]

@@PhysicsNinja thanks sir...

[@dannydeko331]

Every time I thought "why is he doing that??" you explained it perfectly.

[@joudboushi4062]

Wooow , tomorrow is my exam and you really saved my life , appreciate it ❤️❤️❤️❤️

[@PhysicsNinja]

Best of luck!

[@emersonbatzin8341]

This video made the concept so much easier, i get it now

[@isaacgaray6272]

absolutely phenomenal teaching! Wow

[@nirupamam2814]

I understand everything you said because you know how to reach us😊😊👍👍

[@merawithoney]

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?

[@MathPhysicsEngineering]

If you are interested in an in-depth analysis of the geometry and the physics of projectile motion I recommend: kzbin.info/www/bejne/foHImXaLq5uVe7s&ab_channel=Math%2CPhysics%2CEngineering

[@iBen-ry6pj]

Now everything projectile seems like child's play. Dam easy. Thanks a million 🙏

[@KenzoBibble]

Thanks for making it clearer. I was literally crying genuine tears because of how cooked i am.

[@PhysicsNinja]

You got this!

[@victor.novorski]

This helped me on my exam ~thankyou from India

[@RizzyMaeBueno]

Thank you! you make my physics more easier.

[@EwanOkyere]

Thank you now I understand this topic wayyy better

[@christinechoi5756]

Wow u are an amazing professor!! Easily understood

[@PhysicsNinja]

Thank you!

[@aSenseSeeker]

You are the reason I did not get an F in my midterm, thank you Physics Ninja

[@dylaninho2500]

Great vid, in the last equation you can simplify 2 sin theta cos theta to -> sin2theta

[@bcomplexllc-kitchensolutio7248]

Nicely done my brother.

[@PhysicsNinja]

Thank you kindly

[@hannahhiller6059]

I wish you were my teacher. Thank you so so much.

[@PhysicsNinja]

Wow, thank you!

[@lillycrochet]

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 :)

[@donsolellizekristine6776]

Thank you so much! the explanation is very nice and smooth thank you again sir have a great day ahead

[@tropicalermine]

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

[@fanparazzienationfanparazz3360]

True!

[@dennisrayrosas3175]

May i ask how did you simplify that ymax formula to its simplest form from that more complex ymax formula, i am just confused

[@JuiorHenry]

Very clear explanation Thank you 😊

[@PedroHenriquePS00000]

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?

[@julianmccallum8812]

bro thank god for this man

[@blessingsmngomba]

Wow this is great

[@vanessachen2330]

thank you for your effort, patient and very precise explanation.

[@tanmoydev]

Can you make a video on the equation of trajectory of the projectile motion😊

[@PhysicsNinja]

Yes, maybe this weekend

[@tanmoydev]

@@PhysicsNinja thank you sir 🤗.

[@donabhadra-3016]

you are simply excellent, this video literally cleared everything...

[@eshiwanishem2964]

nice work but what about the angle at which the particle hits the ground?

[@guapdoctor4534]

I have the max height and range and need to find initial velocity, I’ve been stuck for ages send help

[@PhysicsNinja]

I will post a video tonight on a basketball problem that deals with this. Stay tuned!!

[@guapdoctor4534]

Physics Ninja thank you so much!

[@clarkkimo9105]

Thank you so much. this is very helpful.

[@zulyc8641]

THANK YOU!!!!! it's so clear now, i was so confused. earned a sub

[@AdajiJoy-y3x]

Plus can I have a detailed explaination on the mathematics involved in deriving max height

[@icedoutzey]

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

[@RonaldNaibei]

You really make me proud sir I now know what I did not do

[@PhysicsNinja]

You got this!

[@ノーネームノーライフ]

At 6:28 why above equation before g has 1/2 but below equation doesnt have?

[@jmk-2277]

Thank you sensei, how may I repay you for your help.

[@twangerrrrrr]

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 😭

[@danisaangel]

im in lecture, wit this playing instead , im cooked

[@sandeepsinghbhatti4084]

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

[@PhysicsNinja]

Good luck with the MCAT

[@thristioncupid6883]

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;

[@rinxalee8885]

This helped, thank you.

[@j.gordonleishman6401]

2 sin a cos a = sin 2a. Therefore, max range is when a = 45 degrees.

[@yijowee5810]

omg so clear tysm 😊

[@keilacabrera9282]

If the angle is not given, how would you go about solving the problem?

[@wtlau5208]

good teaching thnak you

[@PhysicsNinja]

So nice of you

[@coleflynn7538]

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?

[@coleflynn7538]

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?

[@qwertyui90qwertyui90]

What about the X value when Y is at its max

[@xinyinpianoanimemusic1697]

1/2 of range, since it is symmetrical

[@TsionMoshe]

Thank you teacher❤

[@sunildesilva9606]

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?

[@عبدالإله-ح1د]

you are damn good professor.

[@lightmethods7766]

when its physics exam tomorrow and ur so cooked u play the video with 2x speed

[@christineebdalin6733]

But what if there's no given angle? The only given are distance and time. How to find the maximum height?

[@duttroach8489]

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.

[@beymonbros2785]

@@duttroach8489 what happens if you have no given angle but only have speed and distance/range how would you solve for Vy?

[@nilaypatel5367]

this video was insanely helpful, thank you so much

[@maikrorg7171]

Very helpful!

[@Kvarggistapäivää]

14:40 Ehh, why is V0sin theta squared? As you can see, it is not squared on the velocity formula

[@Kvarggistapäivää]

Can anyone help pls ?

[@xinyinpianoanimemusic1697]

@@Kvarggistapäivää substitue t top into equation (2) , u'll get Vosintheta x Vosintheta/g ...

[@ModestusHaundapiti]

Very interesting 🎉

[@joshuasanchez7793]

What if the object is initially displaced upwards?

[@masterboijoe6687]

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?

[@PhysicsNinja]

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.

[@MathPhysicsEngineering]

If you are interested in an in-depth analysis of the geometry and the physics of projectile motion I recommend: kzbin.info/www/bejne/foHImXaLq5uVe7s&ab_channel=Math%2CPhysics%2CEngineering

[@anilkumarsharma8901]

which quadratic equation satisfy this path ???

[@nurhanimhasbullah8176]

for time up, the formula is velocity x divide by -9.8) , so the time will become -ve value. is it true?

[@junex147]

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?

[@PhysicsNinja]

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

[@eshanisandali8726]

Sir can you help me. I nedd a some biomechanics problems.( segmemt lenth segment mass , COM )

[@mshafimir-pl7xj]

sir I think v is not initial Velocity v is final Velocity and u is intial velocity...

[@hadrizharif]

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?

[@dinukaherath7155]

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.

[@yeetman9k867]

Thanks im using this for a program on a rocket

[@princessdheannsabanal6297]

Hi can I just ask how to solve time in projectile motion? The t=1s t=2s t=3s

[@DonSimone1996]

You're an iron man, wow, that's great. Huge accomplishment. That jacket is cool.

[@PhysicsNinja]

There’s no greater accomplishment in life than finishing an Ironman.

[@DonSimone1996]

@@PhysicsNinja you're damn right, sir

[@pejanaamieljesusb.5865]

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

[@nood1le]

Awesome video

[@mmjxsn]

I was told that range was |v|^2 * sin(2theta) / g, is this the same as 2Vo^2 * sin(theta) * cos(theta) / g?

[@PhysicsNinja]

Same, using a trigonometric metric identity 2sinxcosx=sin2x

[@fademusic1980]

How come i now understand ballistic calculations but i still cant understand polar patterns being taught by my precal II teacher

[@alinesamara9607]

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

[@nwto235]

Thanks a lot, sir.

[@lidiauni6057]

thank you so much!!!

[@PhysicsNinja]

You're welcome!

[@theengineer9910]

THANK YOU

[@akber_4]

شكرا جزيلا لك شرح ممتاز جدا

[@iamvan7243]

What would happen if the initial velocity upon launch is 0? Or is that not possible?

[@PhysicsNinja]

That’s the easiest case- Maximum height=0, Range=0

[@iamvan7243]

Thank you! i get it now

[@dialafakhrddine494]

This is 10/10

[@JhayraldBenitez]

can total delta y displacement can be negative?

[@PhysicsNinja]

Yes, if you drop inside a hole. Negative displacement simple tell you the direction. Negative would down relative to where you started.

[@JhayraldBenitez]

@@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

[@invisibleunited941]

Sir detail may parai

[@Anonimo-ew5eb]

what if I have to find the time when the y displacement is 10 ?? considering there will be two answers

[@duttroach8489]

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.

[@AdajiJoy-y3x]

why is a = -g when g is directed downwards

[@PhysicsNinja]

Down is taken to be the negative direction.

[@michelleeo4100]

G is pointing down so it negative but where’s the 1/2 gt square coming from

[@anleal9587]

the kinematic equation

[@thegreatestfamily6398]

i want to say only let GOD bless you

[@PhysicsNinja]

Thank you!

[@niceone1814]

is dy the maximum height?

[@tadeusalweendo]

CALCULATE THE VELOCITY WITH WHICH THE BALL LEAVESTHE PLAYER HAND IN 5 SECOND? WHEREBY NO DISPLACEMENT. HELP PLEASE