Let me show you an application of the quadratic equation that might be used in physics.
Quadratic equations are when you have things like this, the equation that let's you solve these things is this x
equals minus b plus or minus the square root of b squared minus 4ac all over 2a.
Equations of this sort pop up in many different applications, and I want to show you one from physics
that relates to something called projectile motion.
First thing I'm going to do is write down the equation that would be solved in this case.
This equation looks pretty complicated right now, it has a lot of things in there that you're not going to be familiar
with that don't look so recognizable, but I want to simplify this a bit.
First of all the y, the y with a little subscripted zero on it,
the v with the subscripted zero and a y on it and the g will all
be numbers, and so let's me write down some typical numbers that might show up in something like this.
Here are some typical numbers, y equals zero, y with the sub-scripted zero equals 11.2,
the v with a sub-scripted zero y: 21.5 and g is 9.81.
Now here's what all of this might have to do with.
Projectile motion means the motion of something that's thrown, or launched, or fired, or something like that on
earth, or it could be on another planet, but it's something that is under the influence of gravity
and it's going to fall toward the ground.
So you might pretend that you have some kind of a cliff, or a building that's what this kind of step thing
represents and you have something that's just launched in that direction.
What this represents right here, I guess I should put some arrows on it or something, is that y with the
subscript of zero, and what that represents is the initial height of the thing that it thrown, or launched, or fired.
So that's the y sub-0, and here this 11.2 would be measured in meters, I'm breaking a rule by not writing
down all of the units, although, I can stick them here.
Okay, the v with the subscript of 0y on there would represent the initial speed in the y direction, and this
would actually be in meters per second.
A person could throw a rock, or a baseball at that speed with an initial speed in the y direction.
The little g over here, is the acceleration due to gravity on earth. When you drop a rock or something like that on
earth; if you ignore air resistance it will have this acceleration and we have it in meters per second
squared, don't worry about that too much, we're just going to deal with the numbers. And this equation
right here, if I can get a box drawn around it, is the equation that describes this motion, and it has a minus
sign on that last term because things fall downward and we usually let upward be the positive direction.
Now I've got a y equals zero here, we're going to be figuring out when
does the thing drop down to the ground.
Okay, so y equals zero there,
now there's something different from this about the quadratic equations you've solved
before; the equations you've solved have all had x as the variable.
In a projectile motion problem the variable is t, which represents time.
Let me rewrite our equation with these numbers plugged into it.
Just by replacing variables with numbers this equation probably looks less frightening then it did before.
Okay, I've simplified it further here, I have 11.2 plus 21.5t minus 4.91t squared
and that's the equation that I want to solve.
Way over here I've got that, that whole thing is equal to zero.
So here is what I have, zero equals 11.2 plus 21.5t minus 4.91t squared
and those would be typical numbers.
They're not nice integers like you might see in a math class,
but we can still use that formula to solve this thing.
When you solve the quadratic equation ax squared plus bx plus c equals zero,
the thing that multiplies the squared
term is a, well the squared term is a t squared here. So this is the a that I'm interested in.
I'm underlining that minus sign too so I don't lose it. The thing that multiplies just the plain old x,
well here it will be the thing that multiplies the plain old t;
that's the b and then, there's just a number part.
Here is my number part, so that's how I'm going to apply this.
Now in this case, it's not going to be x that equals something, it will be t that equals something because
my variable is t and not x.
Okay, I'll have a minus b well b is 21.5, so I'll have minus 21.5 plus or minus that square root of
four times minus 4.91 times c, 11.2 all over 2a, which is two times.
Alright, I've plugged in all my numbers here. I can simplify this a bit before I actually plug this into
a calculator and I'll show you how I do that here.
In the denominator, two times minus 4.91 turns it back into the minus 9.81,
so that'll simplify things a little bit.
21.5 squared I can leave alone for the time being, I've got a minus times a minus that turns that part into a
plus. So I'll have a plus four times 4.91 times 11.2, and so there's a little bit of simplification.
Now something I'm going to do right here is calculate everything under this square root and I'm just doing this
step wise a little bit, I find that if I try to do everything at once I tend to make mistakes.
When I calculated everything under the square root I got 682.
So if I go one small step farther with simplifying I'll get t equals minus 21.5 plus or minus, the square root of 682
happens to be... and that will be over minus 9.81.
Now it's just a matter of evaluating this thing, I'll get two answers.
One will be minus 21.5 minus 26.1 all over minus 9.81, when you evaluate this be careful to do the entire
numerator, maybe wrap it in parenthesis before you divide it by the bottom.
So you get the order of operations correct, what I get, 4.85 for that first one, comes out positive.
When I do the plus sign that's on there, I get minus 0.47 for that.
So I get two possible answers on this thing,
and this is an example of solving a quadratic equation like you might
see in a physics class.
This number right here would be how long the rock or whatever it is that you threw happened to be in the air.
