Two-dimensional projectile motion
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Visualizing Vectors in 2 Dimensions
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Projectile at an Angle
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Different Way to Determine Time in Air
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Launching and Landing on Different Elevations
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Total Displacement for Projectile
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Total Final Velocity for Projectile
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Correction to Total Final Velocity for Projectile
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Projectile on an Incline
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Unit Vectors and Engineering Notation
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Clearing the Green Monster at Fenway
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Green Monster at Fenway Part 2
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Unit Vector Notation
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Unit Vector Notation (part 2)
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Projectile Motion with Ordered Set Notation
Launching and Landing on Different Elevations More complicated example involving launching and landing at different elevations
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- Let's do a slightly more complicated 2D projectile motion problem now
- So in this situation I am going to launch the projectile off of a platform
- And then it is going to land on another platform
- It's coming out of the cannon. Let me do it this way just to make it 100% clear
- So this angle right over here is 53 degrees
- And we are going to come out the muzzle of the cannon with a velocity of 90 m/s
- Just to give ourselves a sense of how high it's being launched from
- From the muzzle of the cannon down to here, so this height right over here is 25 meters
- and let's say that this height right over here is 9 m
- And so we are essentially launching this from a height of 25 meters
- I know in the last video even though I drew the cannon like this
- we assumed that it was being launched it from a altitude of 0
- and then landing back at an altitude of 0
- Here we are assuming we're launching it from an altitude of 25 meters
- That's when it's leaving the muzzle
- and it's gonna start decelerating at least in the vertical direction
- as soon as it leaves the muzzle
- and then we're assuming it's not gonna land back at the same altitude
- It's gonna land at a different altitude
- So how do we think about this problem?
- So the first thing you always wanna do
- is divide your velocity vector into its horizontal and vertical components
- You use the vertical component to figure out how long it will stay in the air
- And then you use the horizontal component to figure out
- given how long it's in the air, how far does it travel?
- And once again, we're gonna assume the air resistance is negligible
- So just based on what we did in the last video
- I'll go through all of these steps in this one as well
- If we draw our vector
- the length here is going to be 90
- The angle over here--this is our velocity vector--the angle over here between
- And let me draw the horizontal component. It would look like this
- And the vertical component would look like this
- And so the vertical component of the vector, what would be the side right over here?
- Well, this is the opposite side
- We know from basic trigonometry sine of an angle is opposite over the hypotenuse
- is equal to the magnitude of the vertical velocity--
- I write that subscript Y, for the Y direction. That's the vertical direction
- Over the length of the hypotenuse, over the magnitude of our original vector
- Or if we multiply both sides by 90
- we get that the magnitude of that side
- Now if we want to do the horizontal component
- the horizontal side is adjacent to this
- Soh-cah-toa, cos is adjacent over hypotenuse
- So the horizontal component of our velocity, I'll say x direction
- over the hypotenuse, over 90
- Cosine is adjacent over hypotenuse. Adjacent that's this length over 90
- Multiply both sides by 90, you get that the horizontal component
- Now how do we figure out how long the thing takes in the air?
- We'll use the vertical component for that
- Especially since we're dealing with different levels, we can't use that
- more basic reasoning that, "Hey, whatever velocity we start off at
- it's gonna be the same magnitude of velocity but the opposite direction
- because we're not going to the same elevation
- What we could do is we can use the formula that we derived in the previous video
- that the displacement--let me just copy and paste this a little bit lower
- Copy it and paste it. I'll get it right over here
- So we can use this
- We know that the displacement is equal to the initial velocity
- and we're dealing with the vertical direction right here, times the change in time
- plus the acceleration times the change in time squared divided by two
- So how do we use this to figure out how long we're in the air?
- So what is the displacement if we're starting at 25 m high and we're going to 9 m high?
- So over the course, while this thing is traveling it will be displaced downwards 16 m
- Or another way to think about it is
- our displacement in the vertical direction is going to be equal to -16 m
- Right? Because 25-9 is 16
- And so we can put that into the formula we derive in the previous video. We get
- -16 -- I won't write the units here for the sake of space and simplicity--
- is equal to the initial velocity--we're dealing with just the vertical dimension here
- Remember, it's negative because our displacement is going to be downwards
- We're losing altitude
- Times our change in time
- [plus] the acceleration due to the force of gravity for objects in free fall
- It's -9.8 m/s squared
- but we're dividing that by two. So we have -4.9 m/s^2
- times delta t squared, times our change in time squared
- So how do we solve something like this? You can't just factor out a t and solve it
- You might recognize that this is a quadratic equation right over here
- The way you solve quadratic equations is
- you get everything on one side of this equation and then
- you either factor it out, or more likely in the situation, you will use the quadratic formula
- which we proved in other videos, hopefully, giving you the intuition for it
- to actually solve for the times where your displacement in the vertical direction is -16 m
- I will get 2 solutions here, and one solution will be a negative change in time
- There some time in the past, you're also at -16 m
- That's nonsensical for this problem
- So we want to take the positive value here
- So let's put all of this on one side of the equation
- Let's add 16 to both sides
- On the left-hand side you get a zero
- Zero is equal to--I'll write it in the traditional way that we're used to seeing
- I'll write the highest degree term first. So -4.9 times delta t squared
- and then +16
- All of this is equal to zero
- And this once against is just the quadratic equation. We can find its roots
- and the roots will be in terms of delta t. We can solve for delta t using the quadratic formula
- So we get delta t--
- if this is very unfamiliar to you, review the videos on Khan Academy algebra playlist
- on the quadratic formula. If you don't know where it came from, it also proved it for you
- So it's equal to negative b--b is this right here, the coefficient on the delta t
- I'll write the quadratic formula for those of you who can't remember it
- So I'm going to solve Ax^2+Bx+C=0
- The roots over here is going to be -B
- These are going to be the X values that satisfy this equation up here
- So that's what I'm doing over here. This is the B value, negative B plus or minus--
- It turns out we only care about the plus one, because that's gonna give us the positive value
- B squared, so it's this quantity squared
- -4 times A which is -4.9, times C which is 16
- This radical all the way over here. All that over 2A
- A is -4.9, so 2A is -9.8
- So now we can get the calculator out to figure out our change in time
- I'm just gonna focus on the positive version of it
- I'll leave it up to you to find the negative version and see if
- that'll give you a negative value for change in time
- And that's nonsensical. So we only care about the positive change in time where we get to
- a placement of -16 meters
- Let's get the calculator out
- So we get--let me do this carefully. We have
- these two negatives cancel out. So it's plus 4 times positive 4.9 times 16
- and that closes off our entire radical
- And so this will give me the numerator up here
- And I want to divide that by -9.8
- Oh, I just realized that I made a mistake
- I said that the positive version would give you the positive time
- but now we realize that is wrong because when I took the positive version up here
- I get a positive 2.14 for the numerator, but then we divide it by -9.8
- we'll get a negative value. So that's not gonna be the time we care about
- So we care about the time where this is a negative value
- So let me reenter that. Let me do the negative value
- Let me go back a little bit
- And then let me replace this with a -
- So we look at the negative value, because I want the positive time
- And so now my numerator here is a negative value. So this is actual what we care about
- The numerator is a negative value, you divide it by -9.8
- and you get--I'll just round--14.89 seconds
- So delta t the positive version is equal to 14.89 second
- So my initial comment about wanting to using the positive version was wrong
- because we have a negative denominator, so you want the numerator to be
- negative, and only the numerator is negative, will the whole expression be positive
- So we've got this positive time of 14.89 second
- Let me solve for the horizontal displacement, although this is running long
- So the amount of time that we're in the air is 14.89 seconds
- So if I were to ask you the horizontal displacement
- it's going to be the amount of time we're in the air
- times your constant horizontal velocity
- We already figured it out, our constant horizontal velocity
- So if you want to figure out how far along the x-axis we get displaced
- we just take this time times--that just means our previous answer--
- and that gives us 806 m. So this displacement right over here is 806 m
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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