Physics 3: Motion in 2-D Projectile Motion (3 of 21) Projectile Upward Angle
ฝัง
- เผยแพร่เมื่อ 12 ก.ย. 2024
- Visit ilectureonline.com for more math and science lectures!
In this video I will show you how to calculate the time in air and distance traveled when an object is thrown upward at an angle off a cliff.
You sir are the greatest physics teacher in the world.
Agreeee
@@Emma-od4qx you are from which country can you tell me
I also agree man
I agree
@@allahonlyfuturedecider1506 the country he is from is none of your damn business.
you're saving lives out here, thank you so much
Finally someone on youtube who uses the same terms in the equations in my textbook. I finally understand how to find time and x just from watching this compared to the month of lecture I've gone to so far
It is from about 7 years ago, I can still feel it in 2021, South Africa, Wits University. You are just a star!!!!!!! I wonder the reasoning of a person who dislike this. Thank you Biezen!!!!!!!!!
Thank you for your comment. Great that our videos are reaching South Africa.
Francisco,
With projectile motion you can separate the horizontal motion from the vertical motion. (They are completely independent).
Thus the vertical motion has no bearing on the horizontal motion
Therefore the displacement (in the x-direction) is 50 m.
You can do this without the quadratic formula, Just break the vertical motion into two parts; time, distance up from the origin and then distance, velocity and time down from the top of the projectiles path. Then use the total time to solve for horizontal distance with horizontal velocity.
Thank you, sir. Learned this stuff in 11th grade Physics around '99-2000, but needed to brush up. I'm a game programmer and I needed to calculate the initial speed (and/or angle) at which to throw a projectile (grenade, etc) so it reaches a spot I pick after a predetermined time in the air. Just thought you might enjoy seeing who's using your lectures aside from students :)
TheGnomeDev Wow, this motivates me to study mechanics a lot!
I wish I had discovered your channel when I took calculus and physics. It would have made a huge difference in my grades and my sanity. Though I passed, I cried a few times (not ashamed to admit it) out of frustration when things started to get complicated and I couldn’t understand the concepts. But I’m happy that this channel exists because I am getting better at physics now. The parts I didn’t understand now become clearer with your videos.
That’s why one day I asked if you had 3 back ups of all your videos lol because I would be extremely sad if something happens and all of your videos are deleted. One of my main goals is to watch every single video of every playlist of every topic you have. It’s like being glued to the TV watching your favorite show but for science :)
Yes, we wish we had such a playlist when we went to school. We were often lost and looking for help. That is part of the reason why we started this channel. Good luck trying to watch all of them. There are over 6,000.
Michel van Biezen With about 6000 videos and an average of 10 minutes per video.... that’s about 1000 hours. Assuming I watch 15 videos every other day, it would take me about 2 years. Not too bad :)
Haha, that was funny!
10 m/sec^2 makes sense to me
I think you are gift of God
This was very useful! Thank you so much for the help!
I was always confused with these types of problems 😢......Thank you for clarifying these concepts! 😊
You're welcome 😊
+Michel van Biezen Your videos are awesome, sir, your teaching method feels like having a chat with a good friend.
You're the greatest teacher. Thank you
Thank you. We are glad you like our videos.
Last minute preparing for a Cambridge interview. Thank you so much for you videos :)
so how did the interview went?
Thank you so much for such comprehensive videos. You're incredible.
Great lecture, keep producing these useful video professor. Thank you.!
You are the best teacher ever
Thank you so much!
Great work DR.
In the step where you fill in the equation, why did you change the acceleration (4.9) to be negative? I took high school physics and I'm taking it again now in college, so I vaguely remember some stuff but your videos have helped me a ton!
acceleration due to gravity is - 9.8 m/sec^2
Thank you so much for these lectures! Helping me a lot to understand the topics clearly.
You're very welcome! Glad you found our videos.
thx my science teacher didn't bother to teach this to us and just gave us the question. That required this formula.
We are glad to help.
Really easy to understand because of your examples, thanks a lot, great videos
can we use g as 10 m/s^2 it is easy to calculate.As we are not allowed to use calculator.I am getting 69.3 m when i use 10 m/s^2.Is it right .....
this was the video I've been searching for for weeks. Thanks a lot genuinely.
Glad you found it. You'll find lot's of other good examples on this channel; there are almost 9000 videos.
You are Awesome.... I wish I had you at School.
Thank you
Thank you so much for your videos, they've helped me a lot!
I had a question about x, though. How come we used 3 sig figs when solving for time, but for x it was only 2?
Thank you!
Best physics videos!
This math prob is actually a bit similar to the one I'm answering right now, I'm very thankful I came across this video. It helped a lot!!!
Glad it helped!
Thank you very much i really respect your great effort i subbed, liked and commented
Great video!!! Helped a lot!
Thanks again Sir for your video!
In which case, do we take time in the air (t air)=2 (time calculated to reach the maximum height)?
When the object returns to the height it started from.
Michel van Biezen Thanks a lot sir!
Followed the exact same process but I never changed the signs when you first come out with the quadratic equation. Ended up with a completely different result since the b becomes negative in the quadratic formula. Was that a necessary step or even a valid one?
+Razuu My mistake. Had an error in calculations as well actually.
how would you know which quantities where A B C in the quadratic formula?
Those are given in the problem. y is the final height. yo is the initial height. Vo is the initial velocity, etc.
Wow, I can't believe I've Never seen this video before..... Well Done Michel!!... just for fun I decided to use your Negative TIME of -2.33 seconds and plugged that into the Y-equation... and sure enough.... the Y value is equal to the 50 Meters that we started at.... Imagine THAT?.. :D
Well, there are over 7000 videos now so it is not surprising that you may have missed one. Yes, that negative time does have a real meaning. Physics is fun.
sir,why the height is not 50m +max. height then to calculate t ?
thanks a lot teacher🌹
but I didn't understand why didn't you calculate the the time of curvature and add it to time we had found??
The time a projectile spends in the air ONLY depends on the vertical component of the motion and does NOT depend on the horizontal component of the motion
Hi, how do I find the angle of release in a similar instance where height of release is 1.347m, velocity at release is 9.347 m/s and horizontal distance thrown is 8.77m
You should invest in a better calculator. One in which you can type out an entire equation and see your entire equation at once. Also one that has a solve for x function and one that has a quadratic equation function. Casio fx 115es plus has all of those and it is (now) a very cheap calculator. A newer version of it is the fx 991ex ($19), a bit more expensive but an amazing calculator. You can even save equations as a QR code and scan with with your phone to get it on your phone. Pretty amazing. Makes me cringe seeing you spend so much time solving an equation with that calculator when you can do it with one of these in 30 seconds. great videos by the way, you're the reason why i have a good grade in physics
When I solve an equation, is it better to input the given values for the variables in the first place or keep the variables until the equation is simplified?
Typically you want to keep the equation in symbol format until the end and then you substitute in numbers. But in the case of the equations of kinematics, it is often better to substitute in the beginning.
@@MichelvanBiezen okay, thank you very much
Well explained - deserves a thumbs up.
Hello Sir, thanks for the video, I greatly appreciate the playlist, but I have a question please. Can u please let me know how would I find the max height for this problem, when a projectile is launched upwards at an angle off a building . Thankyou
Find the y-component of the initial velocity. Then use the equation: V = Vo + gt to find the time it takes to reach the maximum height. Then fine the max height. Or use the equation V^2 = Vo^2 + 2 h g where V= 0 at the maximum height and g = - 9.8 m/sec^2 all velocities are the components in the y-direction
@@MichelvanBiezen thanyou sir . The problem I am facing is the height is 2.4m and the projectile is launched upwards at an angle of 45 degrees. The initial velocity is 12m/s . They want us to find the max height of the projectile in the air. I calculated time in air using the method from the previous videos in this playlist and got 1.98s. I calculated y component of initial velocity and got 8.5m/s. I used the formula V=Vo +gt and got V=11m/s ( it was negative -10.9, but from ur previous videos of calculating the final velocity, I realised that the negative sign is ignored when using Pythagoras so I just put it as positive, sorry I may be wrong , still learning haha). I'm not too sure where to go from here. I tried to use the formula of height which I saw in one of ur other videos, which said (Vo^2 sin^2 theta)/2g, where g is positive 9.8 to avoid confusion. Can u please help me out. And personally, I like to avoid any other equations in these type of problems as it makes it very convenient to just use Y=Yo+Voyt+1/2gt^2 and X=Voxt. I saw these two on this play list and it really made it easy to solve the soccer question shown in one of the videos which I had no clue where to start from. Thanks for help!
@@MichelvanBiezen the question is in my year 12 physics in focus textbook (Australia) and it says:
In a moment of frustration, Jen throws her maths textbook out the window, 2.4m above the ground. She throws it with initial velocity 12m/s at an angle of 45 degrees above the horizontal. Calculate the maximum height the textbook reaches.
@@MichelvanBiezen I have an exam in 1 day, can u plese resond asap Sir? Thanks in advance
@@MichelvanBiezen please sir
Live long sir i am great fan of you
uhh 0:30 doesnt velocity in x direction stay constant all throughout? It does not get smaller and smaller
The velocity in the x-direction does indeed remain constant.
saved me thanxxxx
Thank U very much these videos it is so easy to understand
Thanks a lot, sir! isn't it better to write -1/2gt^2 instead of 1/2gt^2 ? Because g is positive.
When using g in the equations of kinematics, g is considered negative since gravity acts in a negative direction
The bow seems great
In the quadratic formula,remember that b is negative so it is -10 not 10, reconsider that. But anyways thank you so much!!
maxrod98,
In this problem b = -10, thus -b = 10, so it is correct in this problem
Michel van Biezen O yes you are right, sorry, anyways thank you for you help, you teach better than my physics teacher. :P
maxrod98 He does. :)
+Michel van Biezen But that's only because you changed the signs earlier to make -4.9m/s^2 positive? Would it be valid without changing signs?
Thank you sir for your informative and relevant videos.
Everything u taught is understandable for me you are A GREAT TEACHER IN PHYSICS 😊😊😊
You are most welcome
How did this equation happened?
y = initial y + VyT = .5gT62.
how is y = initial y ?
Y = initial y when t = 0.
watch other videos. Finally understood it . Thank you sir
Thank you so much!
You're welcome!
I love you, man.
very helpful thank you
Thank you for explaining this! I have a quick question though, instead of using the equation y = y (v) (t) + (1/2) g(t^2) to solve for time, can we use v = v. + a(t) and solve for t?
San Telmo,
Not in this case, since you don't know what the final velocity will be in the y-direction.
Thus use the y = y initial + yt + ...... equation
Oh ok, thank you!
Your second equation will be the best bet for you to find time in most problems.
That also will be the true when finding displacement.
Dont know if anyone will read this but you actually can use that equation. I used that equation to find the time until the projectile hits the highest point. Because at the highest point v = 0. So I had 0m/s = 10m/s -1/2gt^2 because 10m/s is the velocity in y direction and -1/2 g is the downward accelaration from gravity.
Then you use the "normal" formula to find the time the projectile takes to fall to the ground from the maximum height and add the two. This way you get the total air time without a quadratic equation.
thank you sir
Used this to write a Java function to throw an object. Thanks
SIMPLY AMAZING! subbed
Thanks for the sub!
Thanks a lot!
really helps me out
Elias,
Thanks for the feedback. Please share with your friends and classmates.
Just to check, If i were to find the maximum height of the projectile, would it be 55.1 m? I used the Vf^ = Vi^ + 2aX to find the max height. If theres anything wrong, please tell me how I can fix it and what would have been better to use. Thank you!
That is correct.
how would you do this without knowing the height from where you throw the ball?
In that case they probably provided information such as time to reach the ground of final velocity in the y direction or something like that. Then you use the equations of kinematics to solve for the unknown.
Mr. van Biezen, could you do an example of this same problem where we are given Time and need to solve for Height and obtain the Distance or Range? For example, we have Initial Velocity of 30m/s, Angle Theta of 10deg, Height ?, Range/Distance ?
Eduardo,
There are a number of different examples of projectile motion. If you watch all of those videos, you'll have a good understanding of projectile motion.
Thanks!
thank you so much sir it was much more helpful
Very clear teaching.
Sir in the y direction, will u always take the g to be negative?
The word "always" is not a good word, because there may be exceptions. But in the case of kinematics while using the 3 equations of kinematics, g should be taken to be negative.
the final velocity is equal to 37.5 the moment it hits the ground.
@2:37 why -4.9? Why negative?
The acceleration due to gravity is directed downward.
@@MichelvanBiezen where did u get the 4.9
Sir how did you find 10/ms from 20m/s. Sin30? Is there Anyone who can explain it to me?
sin(30 degrees) = 0.5
nice tie btw！
thats awesome
i just got here and i still wondering why the gravity turns out to be minus? i mean is it affected by the vector direction?
Yes, the force of gravity acts downward, so when it is used as a vector, the direction is negative.
Why did you take that final height as 0 and acc. Negative...???
You can make the ground any height you like, but it is traditional to call it h = 0 at the ground level. The acceleration due to gravity is always negative since it acts downward.
Michel van Biezen the fact is that i find physics very tough.... Thank you so much...
What would be the angle at which the projectile hits the ground
Take the inverse tangent of (Vy/Vx) when it lands.
I think you made a mistake be cause the point of release and and 'h' is not equal
We consider the height of release y = 50 m
What if it isn't
what happen if time is equation give two positive answer ???
That will not happen with these types of problems
@@MichelvanBiezen ok , thanks a lot
Thanks
Welcome
I wish this professor was at Texas Tech University :(
Well damn I'm going to fail my test
answer is 88M sir
Shookran!
ur saving my whole life but i will still probably get a d in physics
During life we often run into obstacles, challenges, and disappointments. The key is to learn from them, redouble our efforts and planning and to try again. (I had to retry plenty of classes when I was a student).
Thank you sir
thank you sir