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
Green Monster at Fenway Part 2 Solving the problem to determine the minimum velocity to hit a ball with to clear the Green Monster
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- When we left off, we were trying to figure out: what did the magnitude of our initial velocity
- need to be for the ball to clear the Green Monster at Fenway park, assuming that the batter hits it at a 45 degree angle,
- and he hits it one meter above home plate, and the Green Monster is 96 meters away and it's 11.3 meters high.
- So, the x-component of the displacement when it's at 96 meters, the y-component of the displacement has
- to be 10.3 meters. Not the 11.3. Because it's already
- starting at one meter high, it just has to get 10.3 more, it just needs to be 10.3 meters higher.
- And we set up the problem, we set up our formula we derived in multiple videos
- on displacement as a function of time and inital velocity
- and our acceleration, and now we're ready to go because we have all of the ingredients.
- So let's apply it. And what's really neat is, we'll essentially be able to do
- both the horizontal and the vertical components at the same time now that we've expressed all of
- our vectors in engineering notation, as a combination of scaled multiples of i and j vectors.
- So, our displacement as a function of time, our displacement as a function of time,
- I'll just write it all in green, is going to be equal to v sub i times t.
- Well, v sub i is all of this business right over here. v sub i is all of this business.
- And if you multiply a vector times a scalar, and t is a scalar, it's just a number,
- you just multiply each, you just scale each of its components by that much.
- So this is going to be equal to square root of 2, lemme, lemme make it...
- So, this part, maybe I should, so this part right over here, this part right over here,
- is going to be square root of 2 over 2 times the magnitude of our initial velocity, times time times
- i, so I'm just multiplying this times time. So both of these terms have to be multiplied times time.
- Both of these component vectors.
- Plus square root of 2 over 2, time v sub i, times time, because we're multiplying it
- by this time right over here, times j.
- So that is v sub i times t over there. Maybe I'll make the t in yellow just so you see
- that this t, I've just distributed this t onto both terms on the, on our initial velocity vector.
- So, this thing expands onto all of this business over here. I'll underline it in the same way.
- It expands into all that stuff over there, and then over here we're scaling the acceleration
- vector, we're essentially multiplying it by t squared over 2. So we multiply every term here
- by t squared over 2. Well, there really isn't an i term, we can just ignore it, so all we have is this
- j term. And we multiply it by t squared over 2. So all of this over here is this vector
- times t squared over 2. So if I multiply this by t squared over 2, I get the divided by 2 part,
- 9.8 divided by 2 is -4.9 and then t squared j.
- So this part right over here gives us this right here. I just multiplied
- t squared over 2 times -9.8 to get the magnitude right over here, and it's going,
- it's in the vertical direction, although this negative tells us that we're going down in the vertical
- direction. Now remember, remember, we need to get, we need to figure out, we need to figure out what
- this velocity needs to be. So we have to have some constraints here because we have
- 2 unknowns here. We have a t and we have a v, but luckily we can set up 2 equations.
- Because we said that our necessary displacement, when we just tip the top of the wall,
- has to be, so I call that the necessary displacement.
- So the displacement necessary has to be 96 in the horizontal direction,
- 96 meters in the horizontal direction, 96 meters in the horizontal direction,
- and 10.3 meters in the vertical direction. So, plus 10.3 meters in the vertical direction.
- So how can we set up 2 equations here? So what's the, what's the horizontal component
- of all of this business up here? Well, the horizontal component of all of this is
- what's multipied by the i vector. I can even write, make it orange right over there.
- So that needs to be equal to this right over here, and what's the vertical component?
- Well, the vertical component is all the stuff that's multiplied by the j vector.
- And actually, we can group them, we could factor out a j vector, I guess is one way to think about it.
- And all of that needs to be equal to this! Needs to be this over here. And so that essentially gives
- us 2 equations and 2 unknowns, and allows us to solve for the necessary v to clear the Green Monster.
- So let's try that out. So this first equation, we get, right over here, square root of 2 over 2,
- times our initial velocity times time, needs to be equal to 96, right? The magnitude
- here has to be equal to the same magnitude here. And then if we talk about all of the j components,
- we could factor out all of the j components, we could factor out a j here, so all of this stuff
- is multiplied by j, all of this stuff is multiplied by j, I essentially just factored it out,
- so this is the magnitude in the vertical direction, and that needs to be equal to 10.3.
- So our second, our second equation with two unknowns is square root of 2 over 2, times the magnitude
- of our initial velocity, times time, minus 4.9 t squared, needs to be equal to 10.3.
- And our goal here is to solve for v sub i. To solve for the magnitude of our initial velocity.
- Well what we could do here is solve for t here, in terms of v sub i,
- the substitute back in here, then solve for that v, that, that necessary initial velocity, or the
- magnitude of it. So, to solve for t here, pretty straight forward, you just divide both sides
- by square root of 2 over 2 v sub i, square root of 3 v sub i, these guys over here cancel out,
- and you get t is equal to 96 over all of this business, which is the same thing as saying
- t is equal to, t is equal to, dividing by square root of 2 over 2 is the same thing as multiplying by
- 2 over the square root of 2, which is really just the square root of 2. So let me just do that.
- This right over here, I'll write it over here, so 96 over the square root of 2 over 2, is equal to
- 96 times 2 over the square root of 2. And 2 divided by the square root of two is just
- the square root of 2. So this simplifies to 96 square roots of 2. 96 square roots of 2 over our initial
- velocity. And now we can take this, and we can substitute it back into this other constraint,
- where every time we see a t, and then we'll have one equation with one unknown.
- We will get, we will get square root of two...I'll do that in blue.
- We will get square root of 2 over 2 times the magnitude of our initial velocity time time.
- We just figured out that time is all of this business over here, so it's times
- 96 times the square root of 2 over v sub i, minus 4.9, times t squared. Minus 4.9 times t squared.
- Well what's t squared? T squared is the same thing as this thing squared, so it's minus 4.9
- times 96 squared, times square root of 2 squared, so that's just 2, all of that over v sub i squared.
- Did I do that right? Yeah, if I square this I get 96 squared, times square root of 2 squared,
- which is just 2, over v sub i squared, and then that, it needs to be equal to, that needs to be equal
- to 10.3. And now we just need to solve for v sub i. It might look daunting, but it's not going to be
- as bad, if we just keep, if we keep our heads down and we focus on the problem at hand, I guess.
- So the first thing to simplify: we have a v sub i in the numerator and one in the denominator,
- these 2 cancel out, and then we have a square root of 2 times a square root of 2,
- square root of 2 times a square root of 2 is just 2, and we have a 2 in the numerator and a 2 in the
- denominator, that cancels out. So this whole first term gets simplified to 96, which is nice. So that
- is 96, and then we have minus all of this business over v sub i squared, so let's figure out what all
- of that business is. Let's just multiply it out, get the calculator. So I have 4.9 times 96 squared,
- times 2, which gives me 90316, which is about right, because this will be about 10000. And so yep, 90316. So minus, minus 90316 divided by v sub i squared, is equal to,
- is equal to 10.3. And now we know that, just to simplify, let's just multiply everything times
- v sub i squared, we get 96 times the initial magnitude of our initial velocity squared, minus
- 90316...alright, that's the whole point behind multiplying everything times this, so this, so it's not
- in the denominator anymore, is equal to 10.3 times v sub i squared. And now we can, let's, let's do a
- couple of interesting things here, let's add, well let me just subtract 10.3 v sub i squared
- from both sides, minus 10.3 v sub i squared, and lets add 90316 to both sides.
- Plus 90316 to both sides, and what do we get? On the left hand side these guys cancel out,
- if I take 96 minus 10.3, 96 minus 10 is 86, so I wanna subtract another .3, so it's 85.7 v sub i squared
- is equal to, on this side, these guys cancel out, is equal to, is equal to 90316. Now I can just divide both sides by
- 85.7, 85.7, and I get v sub i squared, and this is the home stretch, is equal to this. Let me get the
- calculator out again. So it's equal to this quantity divided by 85.7, which is equal to 1053, well let's
- just say 1053, or 1054 if we round. So 1054. This is going to be meters squared per seconds squared.
- We didn't write the units, but then to solve for v sub i we just take the square root of both sides.
- So v sub i is going to be the principal root of this, so let's just take the square root of that.
- The square root of that gives us 32.5. 32.5. 32.5. And we're done! This is going to be in meters per
- second, so those are the units that we've been handling, we've been handling with kilometer--I'm sorry,
- with meters and seconds and all of the rest, and so that's the velocity that the ball, the magnitude
- of that initial velocity! When you get stuck in the math you sometimes forget what we're even doing!
- But if you hit something at 32 meters per second at a 45 degree angle, one meter above home plate,
- one meter above home plate, in Fenway Park in this direction, you will just cross, or you will just hit
- the top of the Green Monster. So if you go any bit faster than that, so if you were to go, if you were
- to go 33 meters per second, and we assume that the air resistance is negligible and it's not slowing
- you down, you will be able to cross the Green Monster. At an optimal angle of 45 degrees.
- If you're angle isn't optimal, you would have to put some more velocity onto that thing. Or put a little
- more magnitude on that thing, I should say. And, just for fun, if you want to, you might want to convert
- this into kilometers per hour or miles per hour. That may give you a more tangible, a more tangible
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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