This right here is a picture of the famous Fenway Park in Boston and what I wanna do is figure out, if
I have a batter who hits a ball, and they hit a ball right when it's 1 meter above home plate and they
hit it at the perfect 45 degree angle, the perfect angle you need to have the optimal carrying distance.
I wanna figure out what the velocity, or how fast they have to hit the ball, or what is that initial
velocity of the ball need to be in order for it to cross the Green Monster, in ordrer for it to be a
home run, so the ball's path is gonna look something like that. And I wanna figure out how fast does
it have to be hit. The Green Monster is about 96 meters away from home plate and its about 11.3 meters
tall. So let's think about this a little bit. Our initial velocity looks like this. We're starting 1
meter above home plate. Its going to be hit at a 45 degree angle at some velocity. This is that vector
right over here. The angle with the horizontal is 45 degrees. The magnitude of this vector right here,
let's call it our initial velocity vector, and the magnitude of it, I'll just call it 'lower case initial velocity', so this isn't a vector right here, this is just its magnitude.
So how can we represent this vector in engineering notation, well we just have to figure out the magnitudes
of its horizontal and vertical components, or it's X and Y components. So the magnitude of its horizontal
component is going to be v sub i, its gonna be this magnitude times the cosine of theta - comes from basic
trigonometry, we go into depth in that in previous videos - so v sub i cosine of theta, thats it's horizontal
component, the magnitude of its horizontal component. And the magnitude of its vertical component, we've
done this many times before as well, is the magnitude of the vector times the sine of theta, and actually
we know what theta is so I don't actually have to write the theta - theta is, we know the angle, theta
is the general Greek letter that you would use for the angle - but we know that the angle is 45 degrees
so I say v sub i times the cosine of 45 degrees and then the vertical component is going to be the magnitude of our vector, v sub i
times sine of 45 degrees. And both cosine of 45 degrees and sine of 45 degrees are both square root of
2 over 2, thats the easiest ones to have in your brain, although you can use a calculator if you don't
remember that. And so if we want to write our vector, our initial velocity vector, and now I'm making explicit its a vector
is equal to the sum of these two vectors, so its the magnitude of the x component is v sub i times square
root of 2 over 2, actually let me write is this way, is equal to square root of 2 over 2 times v sub i,
times the magnitude of our initial velocity, times, and then the direction this is just a magnitude right
now, times, and the direction is in the 'i' direction so we can multiply this times the 'i' unit vector,
I'll do that in orange. The 'i' unit vector has magnitude 1 and its pointed in the positive x direction.
So this is telling us the direction, we're going in the positive x direction. Now let's do the same for the vertical component.
The magnitude of square root of 2 over 2 times v sub i, and its going in the positive vertical direction.
Or we can just multiply this times the 'j' unit vector, this is a vector that has magnitude 1 going in
the positive y direction, and so we're just scaling it by this much to get this vector right over here.
Now what we want to do is think about the displacement that this ball is going to have to go through
in order to clear the Green Monster. So in the x direction its going to be displaced 96 meters, and in
the y direction, the wall is 11.3 meters high, but its already starting off 1 meter high so the wall
let me make it clear the wall and I'm not drawing this to scale - let me do this in a different color
I should do it in green cause its the Green Monster - so the wall is 11.3 meters high but the actual
displacement, let me just draw this as a wall, but the actual displacement doesn't have to be 11.3 meters
it has to be, if we're starting at 1 meter altitude, we have to get 10.3 more meters in altitude, so
the displacement has to be 10.3 meters in the vertical direction. So if we want our displacement vector
right when its crossing the wall, it should be, or lets think about it right when its, if it was just
good enough to hit the top part of the wall, let's think about what that displacement vector would have
to be and we'll solve for that velocity and then any velocity better than that will make it go even further
and faster and higher and all of the rest of the things. So right when its crossing the wall, if we want
it to just skim by or just hit the tip of the wall, our displacement vector, maybe I'll call it 'displacement
necessary', when its 96 meters in the x direction - I just put this 'n' for necessary - when its 96 meters
in the x direction, and I won't write the units now just for simplicity, when its 96 meters in the x
direction it has to have and upward displacement of 10.3 meters in the y direction, so plus 10.3 times
the 'j' unit vector, this is our necessary displacement. So lets just think about displacement as a funtion
of elapsed time and then figure out what the necessary velocity would be to get this displacement at
some point, to get this necessary displacement. So our displacement is a funtion of time, let me write
it over here, our displacement is a funtion of time - instead of writing the delta 't's like I've been
doing in all of the other videos I'll just write time - but this is elapsed time since ball hit. So
you could view it as the change in time since the ball hit as well but not writing the delta over and
over again will just simplify the writing a little bit. So that's going to be equal to, and we've proven
this formula to ourselves multiple times, we derived it in multiple videos, but the gereral formula is,
its equal to your initial velocity times elapsed time, I've used delta t in multiple videos, plus the acceleration vector, and actually your
intial velocity is also a vector, plus your acceleration vector, times time squared over 2 and we've
seen this multiple times but whats neat about this problem is since we've already wrote our stuff in
engineering notation we can just go ahead and straight up and apply this formula that we've already derived
in multiply videos. But before we do that you might say "well whats the acceleration vector going to be if we think about it in two dimensions"
well the acceleration vector is the acceleration due to gravity on an object in free fall near the surface
of the planet, so this is going to be, its not going to have any x component - I can just write zero I
don't have to write that - and its y component is negative 9.8, I won't write the units here for this problem to save space
negative 9.8 meters per second squared in the vertical direction, so if we scale the 'j' vector by negative
9.8 the negative points it down and then we scale it by 9.8 this is the acceleration due to gravity.
So we have, I think, everything we need. What we'll do is we'll set up two constraints and then we can
solve for the magnitude of our initial velocity, and I realize I'm already close to ten minutes,
so now that we've set up the problem I think this is a good pausing time you might want to try it for
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