Total displacement for projectile | Two-dimensional motion | Physics | Khan Academy
HTML-код
- Опубликовано: 5 сен 2024
- Courses on Khan Academy are always 100% free. Start practicing-and saving your progress-now: www.khanacadem...
Reconstructing the total displacement vector for a projectile. Created by Sal Khan.
Watch the next lesson: www.khanacadem...
Missed the previous lesson? www.khanacadem...
Physics on Khan Academy: Physics is the study of the basic principles that govern the physical world around us. We'll start by looking at motion itself. Then, we'll learn about forces, momentum, energy, and other concepts in lots of different physical situations. To get the most out of physics, you'll need a solid understanding of algebra and a basic understanding of trigonometry.
About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.
For free. For everyone. Forever. #YouCanLearnAnything
Subscribe to Khan Academy’s Physics channel: / channel
Subscribe to Khan Academy: www.youtube.co...
The solutions of the cuadratic equations are: t1 = 0.36s and t2 = 5.66s. So for the horizontal distance we get two answers: x1 = 1.87m (distance to reach 10m on the way up - that's why the cliff is 2m away: nearer than 1.87m would create an impact) and x2 = 29.50 m (actual horizontal distance of impact)
The positive radical isn't going to be nonsensical, it will represent the point that it reaches 10m on the way up
Thank you! You're a genius
Yes this is precisely what I was thinking I knew there had to be a point on the way up when it was at 10 so it could gain enough height to clear the ledge then arc around and land again .
Thank you for this, I was going nuts wondering what happened to the time to reach the upwards 10m!
DeltaT1=0.36s; DeltaT2=5.67s!
how to we decide that the vertical displacement is 10m? can't the projectile move even higher?
Here is also a simple explanation of why it didn't hit the wall.
Since we figured out the Vertical displacement we can use the horizontal displacement in order to figure how much it moved along the x-axis if it moves more than 2 meter it means that it passed the wall and didn't hit it.
I hope that makes sense.
Thanks again, my teacher over complicated this. Easy once you know what you're doing.
Its fun watching someone who is as versed in math as Sal is. He speaks the language
KHAN ACADEMY IS AWE-INSPIRING.........
There will be 2 positive values for delta t, since the object will have a displacement of 10m as it goes up
so what happens when you have two values for it?
U choose the larger as the larger delta t will be the one where the particle has reached the platform
I believe that if you were to solve for the angle which you determine to be 50 degrees by using arctan and the lengths you provided, the resulting angle is 78.69 degrees. Therefore, would not the projectile hit the wall/platform before clearing the top?
hey but this has a limitation u cannot figure out the total displacement along the x-direction as we do not know the time after it completed it's flight i.e it may have stumbled a few steps after completing it's flight..what happened to that time?
Why did we take the vertical displacement as 10 shouldn't we take the greatest height of the particle
hey sal... how would i calculate the total path traversed by our imaginary object? in other words.how do i calculate the parabolic paths distance?
thanks so much ... extremely hopeful
This video should have more views Physics is just awesome.
If we included all the decimals, would it make sense to end up with 6.344621943 as a time interval?
I had the same problem as you until I realized I was calculating the part under the square root wrong. It should be sq rt of [(29.54)^2 - 4(-4.9)(-10)] = sq rt of [872.61 - 196]. I had accidentally done sq rt of [872.61 + 196] instead which gave me your answer. But with the "- 196" the answer is the one he gets in the video.
I think he made a mistake he substituted g as -10 when it should be +10.
you are my fourth favourite khan
1. shah rukh 2. salman 3. aamir 4.... you
my calculator gives me a different number for the numerator. It gives me -53.18 then when I divide by -9.8 I get 5.43.
My equation: -29.54-(sqrt(29.54)^2-4(-4.9)(-16))/2(-4.9)
because you forget your calculator in a different mode
how do you get 4.9 ?
Johnathan Villarreal he divided the acceleration of gravity by 2
interesting, and confusing... These will help when I start physics next year!
Why is g - if its in an upward direction
g will always be negative because of the gravitational pull that acts on the object to bring it down. In this example, on the 10-meter platform, the ball will be brought down there (there on the 10-meter platform) as its final position ending in a stop.
F
This is because if a ball is going in the upward direction, it is going against gravity. Due to which the g is taken as negative.
1. You did not ask anything to solve. You just began solving the X displacement. And after you began solving the total displacement and the angle. All good, but your confusion at 13:25 can confuse your students. 2. You say that solving the abc formula gives a nonsensical answer with a +, but it gives a positive value. You are incorrect in that
Very long way of calculating a slam dunk
my teacher does not now physics i believe
every time I
second!