Projectile motion with ordered set notation Solving the second part to the projectile motion problem (with wind gust) using ordered set vector notation
Projectile motion with ordered set notation
- Welcome back.
- I now want to introduce you really just to a different
- notation for writing vectors, and then we'll do that same
- problem or a slight variation on that problem
- using the new notation.
- This is just to expose you to things, so that you don't get
- confused if your teacher uses a different notation than what
- I've been doing.
- So when we did the unit vectors, we learned that we
- can express a vector as a component of its x- and
- So let's say I had a vector-- let me just pick a random
- vector just to show you.
- So say I had vector a and that equals 2 times the unit vector
- i plus 3 times the unit vector j.
- That's the unit vector notation, and I actually
- looked it up on Wikipedia, and they actually called it the
- engineering notation.
- That's probably why I used it because I am an engineer, or I
- was an engineer before managing money.
- But another way to write this, and I call this the bracket
- notation, or the ordered pair notation, is you could also
- write it like this.
- We have this one bracket.
- That's the x-component, that's the y-component.
- It almost looks like a coordinate pair, but since
- they have the brackets, you know it's a vector.
- But you would draw it the exact same way.
- So given that, let's do that same problem
- that we had just done.
- Hopefully, this make sense to you.
- It's just a different way of writing it.
- Instead of an i and a j, you just write these brackets.
- Instead of a plus, you write a comma.
- Let me clear this.
- I'm going to do a slight variation.
- This was actually the second part of that problem.
- My cousin gave these problems to me.
- They're pretty good, so I figure I'd stick with them.
- So in the old problem, let me draw my coordinate axes again.
- That's the y-axis.
- That's the x-axis.
- So in the old problem, I started off with a ball that
- was 4 feet off the ground.
- So let's say that's 4.
- And I hit it at 120 feet per second at a 30-degree angle.
- So that's a 30-degree angle like that.
- Its' a 30-degree angle to the horizontal.
- And there's a fence 350 feet away that's 30 feet high.
- It's roughly around there.
- That's 30.
- And what we need to do is figure out whether the ball
- can clear the fence.
- We figured out the last time when we used the unit vector
- notation that it doesn't clear the fence.
- But in this problem, or the second part of this problem,
- they said that there's a 5 meter per second
- wind gust to the right.
- So there's a wind gust of 5 meters per second right when I
- hit the ball.
- And you could go into the complications of how much does
- that accelerate the ball?
- Or what's the air resistance of the ball?
- I think for the simplicity of the problem, they're just
- saying that the x-component of the ball's velocity right
- after you hit it increases by 5 meters per second.
- I think that's their point.
- So let's go back and do the problem the exact same way
- that we did it the last time, but we'll
- use a different notation.
- So we can write that equation that I had written before,
- that the position at any given time as a function of t is
- equal to the initial position-- that's an i right
- there-- plus the initial velocity.
- These are all vectors.
- Initial velocity times t plus the acceleration vector over
- 2t squared.
- So what's the initial position?
- And now we're going to use some of our new notation.
- The initial position when I hit the ball, its x-component
- is 0, right?
- It's almost like its coordinate, and they're not
- that different of a notation.
- And then the y-position is 4.
- Easy enough.
- What's its initial velocity?
- Let me do it.
- So we can split it up into the x- and the y-components.
- The y-component is 120 sine of 30 degrees and then the x
- component is 120 cosine of 30 degrees.
- That's just the x-component after I hit it.
- But then they say there's this wind gust so it's
- going to be plus 5.
- I think that's their point when they say that there's
- this wind gust. They say that right when you hit it, for
- some reason in the x-direction, it accelerates a
- little bit by 5 meters per second.
- So the velocity vector.
- This notation actually is better, because it takes less
- space up, and you don't have all these i's and j's and
- pluses confusing everything.
- So the initial velocity vector, what's its
- It's 120 cosine of 30.
- Cosine of 30 is square root of 3/2 times 120 is 60 square
- roots of 3, and then you add 5 to it.
- So what is that?
- Let me just solve it right now.
- So 3 times the square root of 3 times 60 plus 5.
- So let's just round up and make it easier.
- It's 109 meters per second.
- 108.9, so let's just say 109.
- So the x-component of the velocity is 109.
- And the y-component was just 120 times the sine of 30.
- Well, sine of 30 is 1/2, so this is 60.
- Oh, sorry, this should be brackets, although some people
- actually write the parentheses there so it looks just like
- coordinates, but I like to keep it with these brackets so
- that you don't think that these are coordinates since
- you know these are vectors.
- And a position vector is really the same thing as a
- position coordinate.
- But a velocity vector is obviously not a coordinate.
- What's the acceleration vector?
- Well, the acceleration vector, as we said, goes straight--
- that's not straight down.
- This is straight down at minus 32 feet per second squared.
- That's the acceleration of gravity on Earth.
- So the acceleration vector is equal to -- it has no
- x-component and its y-component is minus 32.
- So now let's put these back in that original equation.
- So our position vector, and I'll switch colors to keep
- things from getting monotonous.
- Our position vector-- these are little arrows or one-sided
- arrows-- equals my initial position, and that's 0, 4 plus
- my initial velocity vector, 109, 60 times t, and I'm
- running out of space, plus at squared over two, so t squared
- over 2 times my acceleration vector, 0 minus 32.
- This is actually a little cleaner way of writing it, but
- this is exactly what we did when we did
- it with unit vectors.
- Instead of writing i's and j's, we're just writing the
- numbers in brackets here.
- So let's see if we can simplify this.
- So let me write it in a different color, so that you
- know I'm doing.
- OK, so our position vector t is equal to 0, 4 plus-- and
- now we can distribute this t, multiply it times both of
- these-- plus 109t, 60t plus-- and we can distribute this t
- squared over 2.
- Well, that times 0 is 0.
- And then that times minus 32 is minus 16t squared.
- Now we can add the vectors.
- So the position at any t.
- So let's add all the x-components of the vectors.
- 0, 109t, 0, so we just get 109t.
- And then what's the y-components?
- 4 plus 60t minus 16t squared.
- And there we go.
- We've defined the position vector at a
- function of any time.
- So let's solve the problem.
- Now that they have this wind gust and our x velocity's
- going a little faster, let's see if we can clear the fence.
- So how long does it take to get to 350 feet in the
- Well, this number right here has to equal 350.
- So we have 109t has to be equal to 350.
- And so what's 350 divided by 109?
- 350 divided by 109 is equal to 3.2 seconds.
- t is equal to 3.2 seconds.
- And so what's the height at 3.2 seconds?
- So let's square that.
- 3.2 times 3.2 equals times 16 equals 164.
- So this equals 164.
- And then what's 60 times 3.2?
- 60 times 3.2 is equal to 192.
- So what do we get?
- We get 192 plus 4 minus 164 is equal to 32.
- So our position vector at time 3.2 seconds is equal to 350
- feet in the x-direction and 32 feet in the y-direction, and
- that will clear that 30-foot fence.
- Our ball's going to be two feet above the fence.
- Hope I didn't confuse you too much.
- See you soon.
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