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Projectile Motion with Unit Vectors (part 2)

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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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Projectile Motion with Unit Vectors (part 2)

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