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Welcome back.
Actually, before I teach you how to figure out how high the
ball went-- and you might already be able to figure it
out-- I want to show you really kind of a more
intuitive way for figuring out how fast the ball went.
I used the equation this first time, just to really show you
that this equation can be useful, but I personally
always forget equations, so I find it very useful to have
kind of a common sense way of figuring it out, as well.
And really, this equation was derived by coming up with a
common sense way-- I don't know if that last statement
made sense, but let's move on.
Let's say that same problem, but let's just think about it
without our equations, because that's always a good fall back
when you're panicking in the middle of the exam, and you
can't remember if an equation had a 1/2, or a 2, or a minus,
or a plus, or t, or t squared.
It's good just to think about what's happening.
So when I throw a ball straight up-- I have a, let's
say it's a baseball, it looks like a baseball-- and I throw
it straight up where my velocity initial is equal to--
well, let's say that this the variable.
This is my initial velocity, v sub i.
What's going to happen?
As soon as I throw it up, it's going to start
decellerating, right?
Because I have the force of gravity decelerating it
immediately-- so, with gravity, we're saying minus 10
meters per second squared, right?
This ball is going to keep decelerating until its
velocity goes to 0, right?
The ball, if we were to graph time and distance, where this
is time, and then this is distance.
Right?
At time zero, we're on the ground.
The ball starts off going really fast, then starts
slowing down, then its velocity goes to 0, and then
it starts accelerating in the negative direction, and it
starts going fast, and bam-- it hits the ground again.
What happens is that the ball starts fast, and starts going
slower, slower, slower, slower, and slower, until its
velocity is 0, and then it starts-- you could say
reaccelerating in the opposite direction-- or decellerating,
really-- but reaccelerating in the opposite direction, and
then it hits the ground.
Actually, we know-- assuming nothing about air resistance,
et cetera et cetera-- that the velocity that it hits the
ground with is the same velocity that it left your
hand with, just in the opposite direction.
So there's a couple of interesting things here.
The time at which its velocity is at 0-- so that point right
there-- that's going to be at t equals 2.
We know that this shape is actually a parabola, if you
remember that from algebra two.
Why is that a parabola?
What was the equation for it?
We figured out the equation using that previous formula.
I don't want to use it this time, but what was that
previous formula?
It was change in distance is equal to vit plus at
squared over 2.
It's a parabola, but I think if you had thought about it,
you would have realized also it's a parabola.
And it points downward because a is negative, so the t
squared term is negative-- that's why it opens to the
downward side.
I think that might make a little sense to you.
So what we could figure out is, if we're given a t, we
could say well, half of that number-- let's say that t
equals 10 seconds.
So we know that in 10 seconds, the ball left my hand, went up
some distance, and then came back down and hit the ground.
What we also know then, though, is that t over 2 at 5
seconds, the ball was essentially stationary for
just a moment-- its velocity had decelerated, decelerated,
decelerated, and hit 0, and then right before it started
reaccelerating again, or reaccelerating downwards, its
velocity was 0 at the time t equals 0.
The fact that the ball decelerated from my initial
velocity to 0 in 5 seconds-- what does that tell us?
Well, we have the very simple equation, you know, change in
velocity is equal to
acceleration times time, right?
You probably knew that before watching any of these videos.
And the change in the acceleration, well, that's
just the final velocity minus the initial velocity is equal
to the acceleration times time.
In this situation, what's the final velocity?
Remember, we're not going to go all the way here-- we're
just figuring out from here to time equals 2, right?
So what's the final velocity?
We're saying that point is where the ball is not going
up, and it's not going down, so its final velocity is 0.
So, 0 minus initial velocity is equal to acceleration--
acceleration is the acceleration of gravity--
minus 10 meters per second squared.
I know it's a little confusing, because I'm using
the same t, but let's say that this time is t sub 0, just to
kind of make sure it's not a variable, it's actual time.
So this is t sub naught over 2, right?
Because the ball is motionless right at the peak of its--
we're not an arc, but right at the peak of its travel.
So, it's acceleration times time, but at this time, the
time is going to be t sub naught over 2.
Once again, the 0 doesn't matter, and we can multiply
both sides times negative 1.
We get plus vi, and we get vi is equal to 10 divided by 2--
5 meters per second squared t sub naught.
Which was exactly what we got in the previous video when we
used this formula.
And I think it makes sense to you -- hopefully, this was
kind of an intuitive way of thinking about what happened.
Before actually I do the distance, I actually want to
graph what's happening, because I think it just dawned
on me that that might be something that will give you
more intuition.
I'm all about giving you intuition so you never forget
this stuff.
So this is, if we were to graph-- that's an ugly looking
axes, but I think you'll get the point-- this is distance,
this is time.
We already said it's going to be like a parabola, like that.
Right?
Where this is t sub naught over 2, this t sub naught.
It launches really fast, then it slows down, and then it's
motionless right here, and then it starts
reaccelerating downwards.
If that's the distance, what does the
velocity graph look like?
The velocity graph, I'll draw right below-- I'll draw it in
another color, just for variety.
That's a bold.
So over-- actually, that's not how I want to draw it.
I have to draw the negative side, too.
So this is time, and then this axis is velocity, so we start
off at a positive velocity, right?
We start off at v sub i, and what's going to happen here is
the velocity decreases at a constant rate, right?
And that rate is just the rate of acceleration.
The velocity decreases until at t sub naught-- let me
switch back to yellow-- at t sub naught-- woops, I'm using
the wrong tool, it actually looked like I was drawing
something-- at t sub naught, the velocity now is
negative vi, right?
Remember we said, when the ball comes back down, it's
going at the same velocity, just at
the opposite direction.
This point right here, which is t sub naught over 2, that
corresponds to this point, right?
Which makes sense, because that's the point at which the
ball has no velocity.
And look, the velocity is 0.
So the ball starts going up really fast, slows down at a
constant rate, and what is the slope of this line?
Well, the slope is just the acceleration, right?
Because velocity is the acceleration times time.
And then it's stationary for just a moment, because its
velocity is 0, and then it starts accelerating-- or you
could say decelerating, or accelerating in the negative
direction-- until the point that it's
going at v sub i down.
And of course, if you were to graph acceleration-- if I were
to graph acceleration over time--
acceleration is constant.
It's right here.
Let me just get a line tool.
Acceleration is just a constant minus 10 meters per
second, so it's going to look like that.
And it's just the slope of this line.
If you know calculus, it'll make sense to you that this
line is the derivative of this line, or this curve.
This line is the derivative of this curve.
And even if you don't know calculus, I think it makes
sense to you that this is the slope of this line.
And just so, if you haven't learned calculus, a derivative
is just to a way of figuring out a slope at any point along
the curve, so it's nothing too fancy.
I'll see you in the next presentation.