In this video I'm going to talk to you
about, um, using the quadratic formula in a
real-life problem.
Um, so we're gonna jump
right into it.
And this, um, is video three in
my, I guess, series of quadratic formula
videos. So.
Um, before we jump into the
problem I want to talk to you real quick
about,
um, the real-life scenarios where you will see it.
Uh, the quadratic, um.
We will see the quadratic equations. Um, anytime we have
an object is dropped equation.
Or an
object is thrown equation,  um, and you've
studied this one, Um, this first one.
This object is dropped one. This h equals
negative 16t plus s.
So if you just took
an object and drop it you would use this
formula, uh, this equation right here to find,
you know, um, I guess how long the ball or
how long the item is in the air.
When you
throw something, and this is kind of new
for this particular, um, example.
When something is thrown,
anytime you throw an object, uh, you can
measure how long that object is in the
air by using the object is thrown model.
Okay?
So let's just talk real quick about it.
The, the h represents the height of the
object. Okay?
Uh, the negative 16 is going to
represent kind of gravity and how fast
it's pulling it towards the ground.
Uh, the t will represent time. Okay?
So we have time
here. Maybe I'll write height over here.
So that's the height of the object.
Uh, the v is the new one. Okay? and that's the, uh,
initial velocity. And that's really
important.
So when you throw something
you initially throw it, um with some
velocity.
Um, or hope you do.
Um, so, that's where that comes in.
This t again still
represents time. Alright? And the s
represents your initial height. Alright?
So, those are kind of what we're gonna
look at in this particular problem.
We're gonna have an equation, a, a, problem
where these things will come into play.
So, uh, that's the object is thrown model.
So the problem I'm going to show you guys
today is this.
Um, we're gonna pretend you
throw a ball from a height of six feet
to a friend who is 40 feet up in the stands.
You throw the ball with an
initial velocity of 49 feet per second.
Write and solve an equation to find how
long the ball takes to reach your friend,
and then find the second root and
explain what it means. Alright?
So, um, I'll just leave, uh, leave it on the screen for a
second
so you can reread it if you'd like.
Uh, but essentially what's
happen here's you have, um, uh, an initial
velocity you threw the ball.
Um, and we're
gonna try to figure out, uh, how long, uh, that
ball is in the air for.
When your friend
catches it. So. Um, let's jump into it.
So, do this real quick here.
So, again. We're
going to use that equation. Uh, the object is
thrown model.
So we have h equals, um, negative
16t squared, plus vt plus s. So let's try
to put in some important stuff here.
So, the height of your friend, um, is 40 feet
above the ground.
So that's what we're
gonna put in for h. We're gonna rewrite
the negative 16.
We're gonna rewrite the t squared because that's what we're
trying to figure out.
that it's in the air.
Um, the velocity.
Now we said we
threw it at, uh, an initial velocity of 49, uh,
feet per second.
And again, we throw it up
so it's got to be plus 49. If you threw
an object down you'd have a negative
something there. Uh, so that's our velocity.
We still have t, uh, plus our initial height
was six feet above the ground.
So this is
the, um, information that we needed to plug in.
The 40 was the height we're trying
to figure out. Because that's where the friends gonna
catch the ball at.
The 49 represents our
initial velocity, and the six represents
our initial height that we threw the
ball at.
So, let's go ahead and try to
get this into our standard quadratic
equation. And this kind of goes back to
that other video, uh, video too that I was
talking about.
Notice that we have a 40 over here.
We can't have a 40 over here.
We have to have a zero there because we're
trying to find out when the y-intercept
is, um, is, what, what the x-intercept is or when that
y is zero. So the first thing you need to do
is you need to subtract this 40 from
both sides. Very important.
So we'll say zero
is equal to negative 16t squared plus 49t.
And then we do have a minus 34 right
there. Alright?
Now we have our a, which is
negative 16. We have our b which is 49,
and we have our c which is negative 34.
We can use the quadratic formula now to
solve this. So.
Let's go ahead and, uh, put those numbers in.
Uh, so we have, negative b
is 49. So we'll say a negative 49 plus or
minus the square root of b squared.
Which in this case would be negative, uh, that's a mistake.
Let me undo that. Undo, undo, undo.
It would be 49. Alright? So we'll
put in the b squared. 49 squared.
Uh, minus  four.
Times a. that's where my negative 16 goes.
Times c, which in this case would be
negative 34. All over negative two times a.
Which is negative 16. Alright?
So, we're
gonna go ahead and simplify this. I think
I'm gonna have to go to a different
slide to show it to you guys. Um, but we'll
start, you know, with the 49 squared.
And, you know, simplify that. Then we'll go to
the, the four times the negative 16
times the negative 34, simplify that. So I'm gonna do
that in a different slide.
So, when you do that, uh, you end up with, you know, you have
your negative 49.
Then you have your plus
or minus. Um, 49 squared, uh, ended up equaling
2,401. And we are multiplying three
negatives together. So we'll do that, and
when you do multiply those three numbers
you get 2,176.
All over two times a, which
is two times negative 16. Or negative 32. Alright.
Let's go ahead and simplify what this
number is here.
Um, which turns out to be a really nice number.
We have negative 49,
plus or minus the square root of 225.
When you subtract those numbers.
Now for
those that know 225, you know that it's a
perfect square. So the square root of
that is twa-uh, 15. So we have negative 49 plus
or minus 15 over negative 32. Alright?
So from here you break it into your two
parts and you'd say, you know, negative 49, plus 15, over 32.
And when you do that you'll get an answer.
And then the other one would be negative 49, minus 15, over,
I should probably make sure I put a
negative sign over there. Negative 32.
So again, you just, you know, I would type
these into calculators. Get the answer.
So the first root is approximately 1.06.
The second root is two.
So we have our two roots right here. Alright?
So, let's make sense of what these
mean.
I think one of the best ways to
make sense of the roots is to actually
look at the graph. Uh, so let's just take a
peek, a little look at that.
Uh, we're going
to take a look at the, uh, the first and, uh, fourth
quadrant here.
So, [clears throat] in this particular example, you release
the ball, um, at a height of six feet and the
ball would travel up, and it would come
back down. Okay?
But because we wanna-to
find those roots and, and actually find out
where this, um, crosses the, um, the x-axis, what
we're gonna do is we're gonna use that,
that height of 40. Where your friend was
standing at. Um, and that's really important.
Because what that does is it takes the
eye graph and it shifts it down on the
y-axis. So, if this was six.
Just imagine
that this, if we shifted down 40 we'd end
up at negative 34 right here.
So our graph would look something like this.
The ball would actually come up, it
would go there, and then it would come
back down. Okay?
So, imagine this. When we
subtracted 40 right here. This line, this
x-axis here. This is the line your friend
is standing on. Alright? So your friend
is at that height right there.
You're
throwing it from down at this negative 34.
Y our friend is at the height of zero.
So as the ball goes up, the 1.06 would be
right there. That's that root.
So if your
friend caught the ball on its path up,
that's what that would represent.
Now, if
your friend misses the ball on its way
up to you, up to him, the ball is going to
go up and it's going to reach a maximum
height somewhere up here. And then it's
going to start to come back down.
This point right here is the point where the
ball would cross back down where your
friend could, I guess, and maybe catch it
again if it wins that would be a two.
If it went straight up and came straight
down and he was able to catch it.
Maybe he's diving for it over here or
something like this on a bleacher or
something like that.
Uh, but it's that point.
It's not where it hits the ground.
But it's that point where the, um, uh, that he
could get, that it would come back down
and be at his, at his height again.
So the,  this is the first opportunity
to catch it, and possibly a second
opportunity to catch it. Depending on the throw.
So. That's what the roots mean in
this.
Uh, one last thing I forgot to mention
is the 1.06 represents
seconds. So it's 1.0 seconds
the ball would be in the air.
Uh, and then
it's two seconds is how, um, um, would be on the
way back down.
Is, is when, uh,  your friend could
catch it the, the second time. So. Thanks.
