The very first thing we're going to have to do
to solve a problem like this, is do what I first mentioned
which is that we need this vector that is in a sort of both,
horizontal and vertical to be split into just horizontal and just vertical
because in the equations of motion, that we just wrote down,
we have a velocity in the x direction and the horizontal direction
and we have a velocity in the vertical direction, just the y direction
but we need to find those if we're going to plug them into our equation.
And usually what you'll be given in the problem is not the horizontal velocity and the vertical velocity,
you'll instead be given the total velocity vector, it's magnitude as well as an angle,
a direction that this thing is being launched at,
a direction as an angle from the ground which we call theta and as this Greek letter you see
that looks like a circle with a horizontal line through it, this is just a Greek letter theta,
and is used for angles very often.
So how are we going to find our vertical and horizontal velocities,
if we just know the magnitude and the angle that our projectiles being launched at.
What we're going to do is treat this like a triangle.
We just go back to trigonometry and this will help us solve it
and if you don't remember your trigonometry too well,
don't worry, we are gonna review it right now, the important aspects at least,
and this will be a good reference slide for that.
What we do is again we think of it as a triangle,
and so in this case, the total velocity vector is going to be the hypotenuse of that triangle
so you see this grey shaded area is our triangle
and we have one angle of this triangle given to us.
What we do is we refer to the two legs of this right triangle,
opposite the hypotenuse, we refer to them via the angle,
so we have a side that is opposite to the angle, is farthest away from it,
and then we have a side adjacent to the angle which is touching the angle,
the side that actually has the angle touching it.
In this case the opposite side will be our vertical velocity
and the adjacent side of this triangle will be our horizontal velocity.
So really this question of splitting the vector into its components,
is exactly the same question as if I know the hypotenuse of the right triangle,
can I somehow find the two legs of that triangle.
Using our trigonometry let's do a quick review so we can see how this goes,
the three trig functions that you might've learned,
and we're gonna try to remember now, are the sine of theta,
it's represented like this with the theta in the parenthesis.
For a triangle like this one the sine of theta is defined to be
the ratio of the length of the opposite side to the length of the hypotenuse.
So sine is equal to opposite over hypotenuse, the cosine of theta is again by definition,
defined to be the length of the adjacent side, divided by the length of the hypotenuse
and so you can see one of these gives you the opposite side
and one of these gives, one of these gives you the adjacent side.
Finally, we have this last function which is defined just by the two legs of the triangle,
the opposite and the adjacent.
So the tangent of theta is defined to be the ratio of opposite to adjacent.
And so you can see from trigonometry, we just simply define this functions
based on ratios of different sides of this triangle and ask where this come from.
One way to remember this, there's a lots of ways people use,
but I'll give you probably, the most prominent one and the most oft-cited one,
to remember which one is sine and which one is cosine, and which one is tangent,
is to use this phrase, SOHCAHTOA. Where you can see here an S-O-H-C-A-H and a T-O-A
and this is just sort of a silly way but somehow it sticks in the head,
to remember which one of this is which and what this means is SOH, S-O-H, is sine is opposite over hypotenuse,
that's what these three letters mean.
CAH, C-A-H is cosine is adjacent over hypotenuse, and then lastly TOA, tangent is opposite over adjacent.
And so with this sort of phrase, SOHCAHTOA, sometimes people use that to help them remember
which one is which, and you can use that if it's a helpful mnemonic for you.
Why are these useful? Let's just take the sine and the cosine for now
because remember we are trying to find the two legs of this triangle,
the lengths of these two legs.
All we have to do is rearrange those first two equations that you see there
by multiplying both sides by the hypotenuse.
If we do that, then we have that the opposite side is equal to the length of the hypotenuse,
which again is given in most problems times the sine of the angle theta.
And the angle theta again is also given in most problems,
and so you can find the length of the opposite side of this triangle.
In other words, your vertical velocity just by taking the hypotenuse
which was given to you and multiplying by the sine of theta.
Similarly, you can find the adjacent side by doing the exact same sort of procedure,
by taking the hypotenuse but instead multiplying by the cosine of theta
which again comes from those equations that we already wrote down
about what these trig functions are defined to be.
As I've said what you can do in terms of an actual problem,
is to recognize that the vertical and horizontal components of your velocity vector
are given by the total velocity times the cosine or the total velocity
times the sine of your angle to find the two components, horizontal and vertical.
One thing that might make these easier to remember
if you're trying to remember which one is opposite and which one is adjacent
without having to resort to a funny phrase as we just introduced.
A quick and easy way especially for an exam to remember this
is to know and to memorize that the sine of zero is zero and that the cosine of zero is one.
So why is that useful? We'll take this picture that we have of this triangle
and imagine the angle going to zero so your triangle sort of becomes flat.
You could picture to yourself that the opposite side or the vertical side
should shrink to zero as your angle flattens.
This is a quick way to remember that you get the sine and the cosine correct.
It's very clear when I say it like that, that the opposite side
must be times the sine of theta because the opposite side must shrink to zero as I make my angle zero.
This is an overview slide for what projectile motion looks like.
We have our vertical, sorry our velocity which is pointing off at a particular angle
and then we found the horizontal and vertical components of our velocity.
Now that we've found these two components with the equations of motion that we've written,
just applying the normal equations of motion in one dimension,
we have them applied in the horizontal direction and the vertical direction
and plug in the values that we've just found, we can solve these equations.
To do a quick review of the most important things about the projectile motion
scenario that we just discussed. First, if we're not gonna consider any friction
or any air resistance for the object that's moving through the air, there's no horizontal acceleration.
There's nothing to push it left or right just up and down.
In the up and the down direction or in the vertical direction,
we have a vertical acceleration that is downwards.
And at the surface of the earth, as we've discussed,
this value will be minus G, or 9.8 meters per second squared acting downwards.
We found the components of the velocity as well
and it's very important to remember how to do that
and to be able to do that in any situation.
We found the horizontal velocity is equal to the total velocity,
times the cosine of whatever angle you're given.
And we found the vertical velocity, the velocity in the y direction,
is the total velocity times the sine of whatever angle you've been given.
So remember to have these prepare
and have these ready to whip out, to use for any problem
where you need to rectify vectors into different components.