Projectile Motion: Finding the Maximum Height and the Range

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Күн бұрын

Physics Ninja looks at the kinematics of projectile motion. I calculate the maximum height and the range of the projectile motion.

Пікірлер: 181

@metiburussie8973 3 жыл бұрын

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

@PhysicsNinja 3 жыл бұрын

Thank you so much

@veronicanjovu8353 4 ай бұрын

Well explained thanks for your time

@hael3516 3 ай бұрын

professors as in university... im in highschool learning this 😰

@lacertus2753 4 жыл бұрын

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

@aldrinjames2565 3 жыл бұрын

@CallmeMizami 3 ай бұрын

@@aldrinjames2565Fsfs

@MathPhysicsEngineering 2 жыл бұрын

If you are interested in an in-depth analysis of the geometry and the physics of projectile motion I recommend: kzfaq.info/get/bejne/fraVm6aG25zGdY0.html&ab_channel=Math%2CPhysics%2CEngineering

@torilecky488 Жыл бұрын

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

@PhysicsNinja Жыл бұрын

Glad you think so!

@jessamaecabizares6598 3 жыл бұрын

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!

@lukaround Жыл бұрын

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

@lillycrochet 3 жыл бұрын

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 :)

@burgamin5467 2 ай бұрын

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

@dannydeko331 3 жыл бұрын

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

@iRobotray 7 ай бұрын

This is the best projectile motion video of all time

@donsolellizekristine6776 2 жыл бұрын

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

@vanessachen2330 3 жыл бұрын

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

@ayushagrawal4275 3 жыл бұрын

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 3 жыл бұрын

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

@ayushagrawal4275 3 жыл бұрын

@@PhysicsNinja thanks sir...

@emersonbatzin8341 2 жыл бұрын

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

@HAL_HLA 8 ай бұрын

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

@nirupamam2814 5 жыл бұрын

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

@lilac4196 Жыл бұрын

Such a lifesaver! Better than my professors

@nwto235 4 жыл бұрын

Thanks a lot, sir.

@iBen-ry6pj 8 ай бұрын

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

@dylaninho2500 4 жыл бұрын

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

@aSenseSeeker 4 жыл бұрын

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

@christinechoi5756 3 жыл бұрын

Wow u are an amazing professor!! Easily understood

@PhysicsNinja 3 жыл бұрын

Thank you!

@clarkkimo9105 2 жыл бұрын

Thank you so much. this is very helpful.

@rinxalee8885 3 жыл бұрын

This helped, thank you.

@zulyc8641 4 жыл бұрын

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

@EwanOkyere 6 ай бұрын

Thank you now I understand this topic wayyy better

@nilaypatel5367 4 жыл бұрын

this video was insanely helpful, thank you so much

@julianmccallum8812 10 ай бұрын

bro thank god for this man

@nood1le 3 жыл бұрын

Awesome video

@merawithoney 3 жыл бұрын

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 2 жыл бұрын

If you are interested in an in-depth analysis of the geometry and the physics of projectile motion I recommend: kzfaq.info/get/bejne/fraVm6aG25zGdY0.html&ab_channel=Math%2CPhysics%2CEngineering

@user-uh7rr9vt1v 3 ай бұрын

Very clear explanation Thank you 😊

@user-fw8bl2bh4b 4 ай бұрын

Thank you! you make my physics more easier.

@maikrorg7171 3 жыл бұрын

Very helpful!

@johnmwebela2226 Жыл бұрын

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

@PhysicsNinja Жыл бұрын

I hope you succeed in your studies

@victor.novorski 2 жыл бұрын

This helped me on my exam ~thankyou from India

@tropicalermine 2 жыл бұрын

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

@joudboushi4062 Жыл бұрын

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

@PhysicsNinja Жыл бұрын

Best of luck!

@yijowee5810 6 ай бұрын

omg so clear tysm 😊

@TsionMoshe 3 ай бұрын

Thank you teacher❤

@user-og1yd6le9p 2 жыл бұрын

you are damn good professor.

@jmk-2277 3 жыл бұрын

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

@dennisrayrosas3175 3 жыл бұрын

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

@eshiwanishem2964 3 жыл бұрын

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

@sandeepsinghbhatti4084 3 жыл бұрын

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 3 жыл бұрын

Good luck with the MCAT

@marinamaged962 4 жыл бұрын

Legend

@nirupamam2814 5 жыл бұрын

Very nice

@xxsweatxx6945 3 ай бұрын

If i still dont get it im quiting school

@aigerakabane1312 5 күн бұрын

Soo what happened?

@HusaynTechOfficialChannel 2 жыл бұрын

Thank you 😊

@sunildesilva9606 3 жыл бұрын

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?

@nurhanimhasbullah8176 3 жыл бұрын

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

@bcomplexllc-kitchensolutio7248 10 ай бұрын

Nicely done my brother.

@PhysicsNinja 10 ай бұрын

Thank you kindly

@ModestusHaundapiti Ай бұрын

Very interesting 🎉

@alinesamara9607 3 жыл бұрын

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

@dialafakhrddine494 3 жыл бұрын

This is 10/10

@pejanaamieljesusb.5865 3 жыл бұрын

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

@tony-bt7rg 4 жыл бұрын

Thanx bro

@masterboijoe6687 3 жыл бұрын

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 3 жыл бұрын

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.

@abrarmansur3666 2 жыл бұрын

life saver :)

@akber_4 2 ай бұрын

شكرا جزيلا لك شرح ممتاز جدا

@lidiauni6057 8 ай бұрын

thank you so much!!!

@PhysicsNinja 8 ай бұрын

You're welcome!

@yeetman9k867 3 жыл бұрын

Thanks im using this for a program on a rocket

@twangerrrrrr Жыл бұрын

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 😭

@user-mh5oq4vl4d 9 ай бұрын

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

@PhysicsNinja 9 ай бұрын

You got this!

@jamiyahajibushra1344 6 ай бұрын

Ur the best tnx

@guapdoctor4534 3 жыл бұрын

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

@PhysicsNinja 3 жыл бұрын

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

@guapdoctor4534 3 жыл бұрын

Physics Ninja thank you so much!

@j.gordonleishman6401 10 ай бұрын

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

@anilkumarsharma8901 Жыл бұрын

which quadratic equation satisfy this path ???

@thogoulia4735 3 жыл бұрын

It s very clear

@joshuasanchez7793 Жыл бұрын

What if the object is initially displaced upwards?

@abelalford2935 3 жыл бұрын

Helpful

@thristioncupid6883 10 ай бұрын

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;

@beatricekatongo5871 3 жыл бұрын

Lovely

@user-xf9rq9jw6f 3 жыл бұрын

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

@keilacabrera9282 Жыл бұрын

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

@DonSimone1996 3 жыл бұрын

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

@PhysicsNinja 3 жыл бұрын

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

@DonSimone1996 3 жыл бұрын

@@PhysicsNinja you're damn right, sir

@hadrizharif 2 жыл бұрын

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 2 жыл бұрын

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.

@wtlau5208 Ай бұрын

good teaching thnak you

@PhysicsNinja Ай бұрын

So nice of you

@junex147 2 жыл бұрын

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 2 жыл бұрын

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

@princessdheannsabanal6297 Жыл бұрын

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

@etonefelix Жыл бұрын

Great

@Anonimo-ew5eb 3 жыл бұрын

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

@duttroach8489 3 жыл бұрын

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.

@tanmoydev 11 ай бұрын

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

@PhysicsNinja 11 ай бұрын

Yes, maybe this weekend

@tanmoydev 11 ай бұрын

@@PhysicsNinja thank you sir 🤗.

@eshanisandali8726 10 ай бұрын

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

@cece8309 10 ай бұрын

i heart you.

@icuppu2 3 жыл бұрын

Great explanation, liked and subscribed. I can definitely use your equations when the next extinction asteroid strikes and rips away our atmosphere, or if I go to the moon where there is no air drag. How about a video with air drag of say a BB weighing 0.334 grams and Vo of 204.5 m/s and calculate for Vy and Range; else, it's all theoretical, but not of our reality, but very interesting and I highly appreciate what you have done in such an interesting and entertaining educational manner. Stay safe and God speed.

@iamvan7243 4 жыл бұрын

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

@PhysicsNinja 4 жыл бұрын

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

@iamvan7243 4 жыл бұрын

Thank you! i get it now

@coleflynn7538 2 ай бұрын

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 2 ай бұрын

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 2 жыл бұрын

What about the X value when Y is at its max

@xinyinpianoanimemusic1697 2 жыл бұрын

1/2 of range, since it is symmetrical

@fademusic1980 8 ай бұрын

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

@christineebdalin6733 3 жыл бұрын

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

@duttroach8489 3 жыл бұрын

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 3 жыл бұрын

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

@niceone1814 2 жыл бұрын

is dy the maximum height?

@samiya4798 2 жыл бұрын

10Q keep it up

@mshafimir-pl7xj 3 ай бұрын

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

@taxsportsfanpower8324 3 жыл бұрын

Goood bro

@michelleeo4100 3 жыл бұрын

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

@anleal9587 3 жыл бұрын

the kinematic equation

@user-iz2zo8pv4q 11 ай бұрын

i love you bless you

@mmjxsn 10 ай бұрын

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

@PhysicsNinja 10 ай бұрын

Same, using a trigonometric metric identity 2sinxcosx=sin2x

@user-vb3xl9sx2x 10 ай бұрын

can total delta y displacement can be negative?

@PhysicsNinja 10 ай бұрын

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

@user-vb3xl9sx2x 10 ай бұрын

@@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

@ian_bruh1 3 жыл бұрын

Examples would be really fucking helpful

@bananachipsyum636 3 жыл бұрын

Don’t be mean :/

@user-bh4gw7ck4d 3 жыл бұрын

For anybody needing maximum height you need to square the answer to velocity x (sin) angle and then divide by 2 x 9.8. This video makes a bit of a pigs ear of explaining that