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# Projectile Motion with Unit Vectors : Determining the position vector as a function of time

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- Welcome back.
- We're now going to use our unit vector notation to solve
- up a projectile motion problem.
- We definitely could have solved this without the unit
- vector notation, but what we'll see is, when we do it
- this way, we can do a lot more just kind of not having to
- draw pictures, and we can actually get a little bit more
- intuition into what happens with the problem and solve it
- in a deeper way.
- So let's start with a problem.
- Let's say someone hits a ball that's starting 4
- feet off the ground.
- Let me draw the axes.
- That's my y-axis.
- We're only going to deal with the first quadrant and so
- that's all I'll draw.
- That's y, that's x.
- Let's say the ball starts off 4 feet off the
- ground, so that's 4.
- That's where the ball starts.
- What does it say?
- OK, say you hit a ball that's 4 feet above the ground with
- an initial velocity of 120 feet per second.
- The ball leaves the bat at a 30-degree angle with the
- horizontal and heads towards a 30-foot fence 350 feet from
- home plate.
- OK, so let's see what this.
- so let's draw the velocity vector: 120 feet per second at
- a 30-degree angle.
- And this won't be drawn to scale just because I don't
- have the space to draw it to scale, but the velocity
- vector's going to look something like this.
- And that, of course, is at a 30-degree angle to the
- horizontal, if this is the horizontal.
- What does it say?
- OK, and it heads towards a 30-foot fence, 300 feet from
- home plate.
- So 350 feet is going to be way out here.
- And let's says this line keeps going.
- This is 350 feet.
- It's not drawn to scale because if this 4, 350 would
- be much more than that.
- It's a 30-foot fence.
- Let's say we're in Boston and this is the Green Monster.
- I don't know if that's 30 feet high, but Let's say the fence
- goes up roughly to 30 feet, so that's the fence.
- And the first question they say is does the
- ball clear the fence?
- So let's see if we can do this using vector notation.
- Just so we start off with-- so we don't get too messy with
- this problem, because I think I'll run out of space if I
- did, I'm going to give you an equation, and this equation
- should make sense to you from the projectile motion problems
- that we did earlier in physics without using
- the unit vector notation.
- If it doesn't, I'll maybe make another video where I prove
- this to you.
- But in general, everything we did in the projectile motion
- problems to date, we essentially broke it down into
- two one-dimensional problems and then solved each of the
- dimensions.
- Now we'll keep everything as vectors and
- solve it all at once.
- So this is exciting.
- So let's say p for position.
- So the position at any point in time, and the position is
- going to be a vector.
- The position at any point in time-- and
- this is a good thing.
- I don't want you to memorize it.
- I think it should be intuitive for you and then-- especially
- if you can prove it.
- And I'll prove soon. it to you in the next video, just in
- case it doesn't make a ton of sense, although it should make
- sense from the projectile motion problems. The position
- at any given moment in time is equal to the initial position,
- and that's going to be a vector,
- plus the initial velocity.
- That's a vector-- so I put these funny little arrows on
- top-- times time plus the acceleration-- and, of course,
- acceleration is a vector-- times time squared over 2.
- This is a good thing to know, and it's actually even a
- better thing to be able to prove.
- So let's see what we can do with these.
- And what's interesting here is all of these are now vectors.
- This would have worked if they weren't vectors, if we were in
- one dimensional domain, but what's cool now is we can do
- this in three dimensions, or five dimensions, or whatever
- dimensions.
- So let's see what we know.
- What's the initial position?
- So p i.
- It equals what?
- So how much do we move in the x-direction?
- Or what is the multiple times the i vector?
- Well, we haven't moved at all in the-- this is the initial
- vector, the initial position.
- So it's only 4 in the y-direction, or 4 times the j
- vector, so it has no x-component.
- So 0 times i, times the unit i vector, plus 4 times the unit
- j vector, or the unit y vector, or you can just write
- that as 4j, right?
- You don't have to write this at all.
- What's initial velocity?
- Well, the initial velocity-- so it has magnitude of 120
- feet per second.
- It's at a 30-degree angle.
- So what's its x-component?
- Let me do this in a different color.
- What's the x-component?
- The x-component would be this vector right here.
- Well, it would be 120 times the cosine of 30 degrees.
- And you can review trigonometry or review even
- the early projectile motion problems to get that.
- 120 cosine 30 degrees.
- We'll multiply that times i.
- And then what's the y-component?
- Well, that's just 120 sine of 30 degrees.
- Let me do it in a different color, just for a variety.
- Plus 120 sine of 30 degrees j.
- That's its components in the y-direction.
- And let's see, so we can rewrite the initial velocity.
- What's cosine of 30 degrees?
- It's square root of 3/2.
- Square root of 3/2 times 120.
- That's 60 square roots of 3, and that's times the x unit
- vector, which we denote by i with a little
- funny hat on top.
- I'm going to switch you colors again, just so I
- don't get all messy.
- shouldn't use that color.
- I like to keep it colorful.
- What's sine of 30 degrees?
- Well, that's 1/2.
- 1/2 times 120, so that's 60j.
- OK, so we've figured out what the initial
- position vector is.
- We know what the initial velocity vector is.
- What's the acceleration vector?
- Well, what's the only force acting in this problem after
- we've hit the ball?
- Well, the only force is gravity.
- And since we're dealing with the English system, how fast
- does anything accelerate on Earth with gravity?
- Well, the acceleration vector just pulls down, right?
- This is the acceleration vector.
- It just pulls down at 32 feet per second squared.
- So the acceleration vector once again has no x-component,
- just like the position vector had, 0 times x, and it's
- pulling down, so it's y-component is
- negative minus 32j.
- Or we could just write this as minus 32j.
- We know all of the different vectors in this equation.
- Let's fill them in.
- So now we know.
- Let me switch colors again.
- And this is neat, because now we're going to be able to
- figure out the x- and y-position of the ball at any
- given time t.
- The position as a function of t-- and this is neat.
- This is a vector as a function over time.
- The initial position we just said was just 4j, right?
- It only has a y-component.
- It equals 4j plus the initial velocity times time, so time
- times initial velocity.
- What's the initial velocity?
- 60 square roots of 3i.
- This is t right here.
- 60 square roots of 3i, and then plus 60j-- that's this I
- just filled in-- times time, right?
- Because it's the initial velocity times time.
- Then plus-- I'll take the t squared over 2 out front.
- t squared over 2 times the acceleration vector.
- Well, what's the acceleration vector?
- Well, that's just minus 32j, where j is the unit vector in
- the y-direction.
- And let's simplify it.
- That equals 4j plus-- distribute this time-- 60
- square roots of 3ti plus 60tj.
- Then 1/2 times minus 32 is minus 16.
- So you get minus 16 t squared j.
- Oh, look, I've actually run out of time.
- So I'll continue this in the next video.
- See

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