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
Total Displacement for Projectile Reconstructing the total displacement vector for a projectile
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- Let's do another example where we're projecting something and it lands at a different level
- Let's also figure out some other interesting things. We'll figure out
- what the actual velocity vector is when it's landing
- So both its magnitude and direction
- Let's say we're launching something from ground level
- and we're going to launch it at a pretty step angle over here
- Let's say we launch it with an angle of 80 degrees
- and it is going to be going at 30 meters per second
- So that's the length of the vector; that's the magnitude of that vector
- And let's say we want to make it land on this landing
- And this landing over here has a height of 10 meters
- So what I want to do first of all is figure out how far along the landing do I actually land?
- And maybe I'll add some other information right here
- From the launching point to the beginning of the landing--
- let's say this right over here is 2 meters
- So we just want to know how far along the landing do we land
- So like we did before, we want to break this vector into its horizontal and vertical components
- I'm gonna go a little bit faster in this video
- hopefully we're getting hold of this type of thing
- So the vertical component of our velocity
- is going to be equal to the magnitude of our total velocity 30 m/s
- And the horizontal component of our velocity is going to be
- Again cosine is adjacent over hypotenuse
- I'm gonna skipping steps. In the last two videos I go into this in a much more detail
- So how much time do we spend in the air?
- So once again, in the last few videos, we saw that we can look at our displacement
- If we want to figure out time in the air, we know that displacement is equal to
- the initial velocity times time--
- let me write change in time, that's technically more correct
- plus acceleration times change in time squared over two
- Now in our situation, we know what our initial velocity is
- We're talking about the vertical direction right over here
- So our initial velocity is going to be this
- We're trying to figure out how much time in the air
- and then the vertical component determines that
- because at some point when it hits back to the ground
- it's not going to be traveling anymore, so that's what determines this time in the air
- So we know the acceleration. Remember the convention when we're dealing with the
- vertical dimension is, up is positive, down is negative
- So this is -9.8 m/s squared
- And then what is the total displacement that we're gonna have?
- Well, we're starting at ground level
- And we're just talking about the vertical, remember that
- So our total displacement is going to be 10 m
- So this value right here is going to be 10 m
- Times our change in time
- So this is -9.8 divided by 2, so it's -4.9 m/s squared, times delta t squared
- And then we can subtract 10 from both sides and write this into a
- traditional quadratic equation form
- So we get -4.9 times delta t squared + 29.54 times delta t -10 is equal to zero
- And then we can use the quadratic formula to find the roots of this
- So the delta t's that satisfy this quadratic equation are going to be negative B
- So -29.54
- of 29.54 squared, B squared
- -4 times A which is -4.9, the negative times and negative is positive, so it's +4 times +4.9
- times--I shouldn't have jumped so fast to get rid of the negatives
- So it's gonna be -4 times A which is -4.9, times C which is -10
- So just A times C, -4.9 times -10
- These two guys, their signs are gonna cancel out
- All of that over 2A, over -4.9 times 2, so -9.8
- And like we saw in the last video, we want a positive value for this
- and negative time is nonsensical. That's kind of going into the past
- So we want a positive value
- And since we have a negative in the denominator, we want have a negative value up here
- And if we already have a negative value here and if we subtract from that negative value
- we'll definitely have a negative value here
- Then you divide by a negative value. You'll get a positive value
- So we can really focus on the subtracting the radical
- You can try it out. If you try the positive version
- you'll get a negative value for this entire thing
- You can try that out after this video
- just to verify that that will get a nonsensical answer
- So let's use the negative right over here. So we have
- -29.54 - square root of
- 29.54 squared -4 times (-4.9)ĄÁ(-10) which is 49, times 49
- Actually I should add some parentheses. Okay
- Times 49. So this right over here would give me the numerator if I evaluate this
- We've got a negative value, and I divide that by -9.8
- Gives me 5.67 second
- And you can keep the units in there and make sure the dimensional analysis works
- You'll find that it does
- So our total time in the air is 5.67 seconds
- Now what I want to do, the whole point of this, is to figure out
- how far along the this landing we land
- Well, the horizontal component of our velocity is right over here
- We know that our displacement in the horizontal direction
- will be our velocity in the horizontal direction--
- it's a constant velocity
- So it's the same thing as our average velocity in the horizontal direction
- Times the change in time
- I won't write the unit. This is m/s times s and it'll give us the answer in m
- Gives us 29.53 m
- So our total horizontal traveling displacement is 29.53 m
- It's a vector. That is our horizontal displacement, which is 29.53 m
- Now we've done a lot of deconstructing vectors
- What I'm interesting in this video is to construct a vector
- So we know our horizontal displacement; we also know our vertical displacement
- It's positive 10 m
- So what's our total displacement?
- Let me write this down
- So we have a horizontal displacement of 29.53 m
- and we have a vertical displacement of +10 m
- So what is our total displacement?
- We can use the Pythagorean theorem here
- The square of the magnitude of our total displacement is
- going to be equal to the sum of these two squares
- This is just the Pythagorean theorem
- So let me write it over here. So this is the magnitude of our displacement right over here
- The magnitude of our total displacement squared is going to be equal to 10^2 +29.53^2
- To solve for this, we just take the square root of both sides
- If we just take the square root of both sides
- we will get the magnitude of our total displacement--
- let's get the calculator out to do that
- So the magnitude of our total displacement is
- the square root of 10^2 is just 100 + 29--
- I can use all this information
- I'll use the Ans, which literally means the previous answer which is 29.53
- squared
- Gives us a total displacement of 31.18 m
- Of course it's a vector. This is only the magnitude. We also need the direction
- So one way to specify direction is to give you the angle with the horizontal
- And let's call that angle theta
- And once again we can use our trig functions over here
- We can use pretty much any of the trig functions
- But we know the opposite side is 10; we know the hypotenuse here is 31.18
- So why not use sine? Sine is opposite over hypotenuse
- So we know the sin of theta is going to be equal to 10/31.18
- Or if you want solve for theta
- you take the arcsine or the inverse sine of both sides
- Theta is equal to the inverse sine, or arcsin of 10 / 31.18
- Once again get the calculator out to figure out that value
- I'll take the inverse sine--this is the same thing as arcsin
- This says, give me the angle. When I take its sine, I get this value
- So the inverse sine of 10 divided by our previous answer 31.18
- is equal to--
- This says, give me the angle whose sine is 10/31.18
- Here we've constructed a vector. We took its vertical component and its horizontal component
- and we're able to figure out the total vectors
- This projectile in this situation, its total placement--just to make it clear
- Its path will look something like this
- And we've just calculated its total displacement
- And I realize that when I started this problem, I asked you
- I think I was asking, how far along the platform
- We figured out its total horizontal displacement
- So if you want to know how far along the platform, the platform starts 2 m to the right
- so it's really 27.53 m along the platform is where it lands
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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