
Jerry J.
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the vector of the green monster lessons 1 and 2 at fenway park are way to formula technical and you should spend time on expanding on the interactive diagramming of the actual field view and formula so as not to confuse students that have only average skills in vectors, so show more field specific diagrams. 

Avinash C.
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How did Sal get 9.8 on acceleration? 
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 xcomponent of the displacement when it's at 96 meters, the ycomponent 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 kilometerI'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
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