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# Projectile Motion with Unit Vectors (part 2) : Let's see if the ball can clear the wall.

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- Welcome back.
- Where we left off, we were using this formula up here,
- which should make some intuitive sense to you.
- And I'll prove it if it doesn't.
- We're using this, and it's neat, because it involves
- vectors, instead of just pure variables.
- So there's actually vector variables.
- But we're using this to figure out, I hit a ball that was 4
- feet off the ground with a bat.
- I want to see if I can -- and I know its initial velocity
- and its angle.
- And I want to see if it can clear this fence
- that's 350 feet away.
- And we do that.
- We use this formula.
- And we work through it to get the position of the ball at
- any given time-- which I think is neat-- in two dimensions,
- and we can actually use this in three dimensions, if we
- were in three dimensions.
- So let's go back to where we were.
- We got to this step, and now we can simplify this.
- So the position at any given time.
- I'll write the, let's say the x position.
- So what are the terms that involve i, the unit vector i?
- Well, there's only one, right here.
- Right?
- So that's what tells us the x position of the ball.
- So it's 60 square roots of 3 times i.
- And then what tells us the y position of the ball.
- Oh sorry, I got rid of the t.
- There's a t here.
- Times time.
- It's a t.
- And then what tells us the y position?
- We'll have this positive 4j.
- That was its initial position above the ground.
- And then I have 60 tj, and I have minus 16t squared j.
- So what we could do is, and we write plus -- This is going to
- have multiple terms in it, so let's group all of the things
- that are multiplied times a y unit vector.
- So you say, 4 plus 60t minus 16t squared j.
- And there we have it.
- We now have the position-- the x and the y position of the
- ball at any given time, t.
- And actually, let's look at this.
- Let's see if this makes sense.
- So the x position is just a linear function of time.
- And does that make sense?
- Well, sure, if we remember, the 60 square root of 3, this
- was just the x component of the initial velocity, right?
- That was just this right here.
- And since there's no other force acting in the x
- direction, or the horizontal direction, we stay at that
- constant velocity.
- We don't accelerate or decelerate.
- So it at any given time, the distance that we travel in the
- x direction is just the time times the velocity.
- So that makes a lot of sense.
- And that should make intuitive sense from what we learned
- when we originally did the two-dimensional projectile
- motions without using the unit vector notation.
- Good enough.
- So let's look at the y position.
- So if you look closely, this soon. whole y term here, this
- whole thing, it looks very similar
- to our initial equation.
- And that's because we start 4 feet off the ground.
- And notice we didn't have a term like the 4 at the x
- position because we started at x is equal to 0.
- We just define that we're starting here, where x is
- equal to 0.
- Otherwise, we would have a term here for the x position.
- We start four feet off the ground.
- Our initial y velocity in the y direction is 60, right?
- Because that was 120 times the sine of 30 degrees.
- That's this component.
- That component was 60.
- So our initial vertical velocity was 60 feet per
- second upwards.
- And then our downward acceleration all affected in
- the y direction.
- And you take 32 divided by 2, and you get the 16.
- And it's minus because
- acceleration is acting downwards.
- So what we have here is a very neat way of knowing at any
- given time, I could say, I have exactly 2 seconds, where
- is the ball?
- And actually I could use this.
- I could draw a cartoon.
- I could show the path of the ball.
- Like if my cursor was the ball, I could plug this into a
- computer program, and then the computer program could just
- draw the trajectory of the ball.
- But anyway, let's get back to the problem.
- And let's see if we can solve it.
- So we want to know, is the ball at least
- 30 feet in the air?
- Because that's how high the fence was-- Right?
- We said it's 30 feet in the air-- when it has gone 350
- feet in the x direction?
- When it has gone 350 feet.
- So how long does it take it to go 350
- feet in the x direction?
- Well, this term right here has to be 350 feet.
- So let's solve for time.
- 60 square roots of 3 times time has to be equal to 350.
- Let me get the calculator out.
- Or that time-- I'm running out of space-- time has to equal
- 350 over 60 square roots of 3.
- So let's see what we can do.
- Get the calculator.
- So 3 square root times 60 equals-- and now that's in the
- denominator, so let me invert it-- times 350.
- OK, so let's say, 3.37 seconds.
- So this ball is moving fast. It went 350 feet
- in only 3.37 seconds.
- I guess that make sense.
- When I look at the baseball game, the ball's not in the
- air that long.
- So at 3.37 seconds, how high is the ball?
- Well, we can figure that out by using the y component of
- our position vector.
- So let's do it.
- OK.
- So let me clear this.
- Let you me do each of the terms separately.
- Let me do this hard term first. 16 -- I shouldn't have
- deleted that.
- 3.37 -- This doesn't have a squared button on it, does it?
- No.
- So I'm going to have to do it-- times 3.37 is equal to
- 11.53 and then times 16 equals 181.7.
- So this term here, let me do a different color.
- This is equal to 181.7.
- What's 60 times 3.37?
- Remember that was the time, equal to 202.
- This is equal to 202 and you have the 4.
- 202.2 and you add the 4.
- And then we're going to subtract this number, right?
- Because it was minus 16c squared.
- And you say minus 191.7 equals 14.5 feet.
- So at 3.37 seconds, the y position-- So we can even
- write it like this.
- Let me see, I'm running out of space.
- We could say the position at 3 seconds is equal to 350i.
- Right?
- Its x position is 350.
- And what's its y position?
- Plus, well we just figured it out.
- We substituted 3.3 seconds into the y piece.
- Plus 14.5j.
- Right?
- This is its y component.
- This is going to be 350 feet, and it's going to be at 14.5
- feet in the air, so the ball would have flown like this.
- And it's only 14 and 1/2 feet in the air, so it's not high
- enough to clear the wall.
- It's going to hit the wall and bounce off, and that's out of
- the scope of this project, what happens
- after it bounces off.
- But we know, it's not going to clear the wall.
- So hopefully I haven't confused you too much.
- I know I ran out of space, and this looks a little bit hairy.
- But I think you'll see that what's neat about the vectors
- is that now I can do these problems in an arbitrary
- number of dimensions without having to separately draw the
- x and y vectors.
- And what's especially neat is, I can get a function, a vector
- function, that tells me the position of the ball at any
- given moment in time.
- And that's relatively new.
- I don't think we have actually did that in any of the other
- projectile motion problems.
- So that's it for now.
- And I will do a couple more these problems very soon.
- See

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