What I want to do in this video is
tackle a problem that would be considered pretty difficult
for most first-year physics students.
And you, frankly, probably wouldn't
be expected to solve a problem like this
in most first-year physics class.
Or if you're in an advanced or honors class
you might be expected or it might be a bonus problem.
But it's an interesting type of problem.
Because what we're going to do is,
we're going to launch a projectile on an incline.
So maybe we're on the side of the hill.
So it's a hill.
Let me do it in green.
So let's say we're on the side of the hill like this.
And let's say that we know the inclination of the hill.
The hill's inclination is 30 degrees off the horizontal.
So this is the horizontal right over here.
So that is the inclination of the hill.
And then we're going to launch a projectile
at 10 meters per second.
We're going to launch it at 10 meters per second.
And the angle with the hill is 15 degrees.
So at a 15 degree angle with the hill.
And the reason why this is more difficult
than the traditional projectile motion problems is, well, we
could think about it.
The projectile is going to be launched
and it is going to eventually land at some point on the hill.
But we can't do the simple figure
out how long it's in the air using its vertical velocity
because we don't know what the vertical displacement
for this thing is going to be, unless we
know how far down the hill it lands.
Because the further down the hill
it lands, the higher the vertical displacement.
So we have to think about both the horizontal
and the vertical displacement at the same time.
And as we walk through this, you'll
see how that can be done.
So I guess the first thing that we really would always
want to do whenever you want to try
to solve this type of problem is break up our velocity into both
the horizontal and vertical components.
So the vertical component of our velocity
is going to be the magnitude of our total velocity, 10
meters per second, times-- and be very careful here--
not the sine of 15 degrees, but the sine of the angle
with the horizontal.
So times the sine of 45 degrees.
And I go into it in a lot more detail in previous videos.
For the sake of time, I won't go into it.
But it really just comes from sohcahtoa.
If we were to draw the vertical component,
it would look like this.
This is the angle.
The sine of 45 degrees is equal to the opposite
over the hypotenuse or the hypotenuse times
the sine of 45 degrees is equal to the vertical component.
That's where it's coming from.
Let me get rid of some of the stuff I just
drew just so it makes it a little bit cleaner.
And the horizontal component of our velocity is going to be,
by the same logic, 10 cosine of 45 degrees.
Now let's think about what the horizontal displacement
is going to be.
And I'm just going to go straight to the formula
that we've derived in the last few videos.
The horizontal displacement is going
to be our initial-- sorry.
Let's do the vertical displacement.
The vertical displacement-- I could
have done the horizontal displacement first--
but the vertical displacement is going
to be our initial vertical velocity, which
we know as 10 sine of 45 degrees.
And by the way, we could just solve that right now.
What is the sine of 45 degrees?
Sine of 45 degrees is square root of 2/2.
Cosine of 45 degrees is also square root of 2/2.
So both of these values, 10 times square root of 2/2
is 5 square roots of 2.
So this whole thing right over here
is 5 square roots of 2 meters per second.
That's our vertical velocity.
And our horizontal velocity is also
5 square roots of 2 meters per second.
So that simplifies things a little bit.
But anyway, we were talking about
our vertical displacement.
Our vertical displacement is going
to be our initial vertical velocity, 5 square roots of 2,
times our change in time plus the acceleration.
Well, we know what the acceleration is.
It's negative 9.8 meters per second squared.
So let me write minus 9.8.
I'm not going to write the units here so that we say space.
Times our change in time squared.
All of that over 2.
We derive this in several videos, especially the last few
where we do these two dimensional projectile motions.
So this gives us our displacement
in the y direction.
And I can simplify this a little bit.
Our displacement in the vertical direction
is equal to 5 square roots of 2 times
delta t times-- let me do it in the same-- times delta t change
in time minus 4.9 times our change in time squared.
So we know we have this constraint right over here.
So this gives us our vertical displacement
as a function of time.
Let's think about our horizontal displacement
as a function of time.
Our horizontal displacement is going
to be equal to our horizontal velocity, which
is 5 square roots of 2 times our change in time.
Now what can we do next?
Well, we have to have some relationship
between our horizontal displacement
and our vertical displacement.
And that relationship is going to be
given to us by this incline.
So wherever we land-- let's say this
is where we eventually do end up landing.
Well, let's think about our horizontal
and our vertical displacements and what
their relationship would have to be.
So if this is where we land, then this
would be our-- let me do it in the same colors-- that
right there would be my vertical displacement.
I would move that far up.
And then our horizontal displacement
will be this right over here, will be that length
right over there.
So that is our horizontal displacement.
So what is the relationship between
our vertical displacement and our horizontal displacement?
And we know that this angle right over here is 30 degrees.
So we can use some basic trigonometry.
We have a right triangle.
We know the opposite side from the angle.
We know the adjacent side.
And the trig function that uses the opposite and the adjacent
is the tangent function.
So we get the tangent of 30 degrees
is going to be equal to the magnitude
of our vertical displacement over the magnitude
of our horizontal displacement.
And the tangent of 30 degrees, that's
the same thing as the sine of 30 over--
let me just do it over here.
So the tangent of 30 degrees is the same thing
as the sine of 30 degrees over the cosine of 30 degrees.
Let me do this a little bit neater.
And the sine of 30 degrees is 1/2.
This is equal to 1/2.
And the cosine of 30 degrees is the square root of 3/2.
So this is equal to 1/2 times 2 over the square root of 3,
which is equal to 1 over the square root of 3.
So we get the magnitude of our vertical component
over the magnitude of our horizontal component--
is our horizontal component right over here--
is equal to 1 over the square root of 3.
What's useful about this is it gives us
a relationship between our horizontal
and vertical component, or between our vertical
and our horizontal components.
And we can use this constraint right over here
to then solve for one of these two.
And let me show you how we'll do it.
So let's just explicitly write this.
Well, let's do it this way.
Let us cross multiply here, which is really
the same thing as multiplying both sides by the square root
of 3 and the magnitude of our horizontal component.
We get the square root of 3 times the magnitude.
And both of these are going to be positive.
Well, let me just write it this way.
Times the magnitude of our vertical component
is going to be equal to the magnitude
of our horizontal component.
So right, just like that.
So we now have a relationship between the length of these two
vectors.
And we can use this relationship to substitute back
into the constraints that we already have.
So the second constraint right over here,
let's use this information.
The second constraint says that our horizontal component
of our displacement is equal to 5 square roots
of 2 times our change in time.
Or another way of thinking about it,
if we divide both sides by 5 square roots of 2,
we get our change in time is equal
to the horizontal component of our displacement divided
by 5 square roots of 2.
But we also know that the horizontal component
of our displacement is the square root
of 3 times the vertical component of our displacement.
Here I explicitly wrote the magnitude notation.
When we start dealing with either just the vertical
or the horizontal component, I can just
write it like this, because it's either
going to be a positive or a negative value.
And that specifies both the magnitude and the direction.
So what I'm going to do right over here, and obviously
the way I've drawn it right over here, both of these
are going to be positive values.
It's upwards displacement in the vertical direction,
so that's positive by our convention.
And we're moving to the right.
So that's positive, also, by our convention.
So I can rewrite this over here as being
equal to square root of 3 times our vertical displacement.
And all of that's over 5 square roots of 2.
Now, the whole reason why I did this
is, this expression right here contains this information,
contains the ratio between our vertical displacement
and our horizontal displacement.
And it also contains the information of,
how does the horizontal displacement, how does that
change as a function of time?
So our time needs to be equal to this.
So this is our time as a function
of our vertical displacement now,
not time as a function of our horizontal displacement.
And what we can do is, we can use this constraint
with our original vertical displacement
as a function of time to then solve
for our vertical displacements.
So let's do that.
Let's substitute this business right here
for delta t in this top equation right over here.
So if we do that-- and I'll write it big--
we get our vertical displacement, right over there,
is equal to 5 square roots of 2 times delta t.
So it's 5 square roots of 2.
Delta t is all of this business over here.
So 5 square roots of 2 times delta t.
Delta t is square root of 3 times our vertical component.
All of that, really the magnitude
of our vertical component, over 5 square roots of 2.
So that's that right there.
Or actually, we could look at this one right over here.
We could use this constraint.
This is just simplified.
So then we have minus 4.9 times delta t squared.
So delta t squared is this quantity squared.
I'll just write it out.
I don't want to skip too many steps.
So delta t once again is the square root of 3 times
the vertical component.
All of that over 5 square roots of 2 squared.
And now what does this give us?
So now we literally have a quadratic equation
with only one variable.
So we can solve for this.
But let me rewrite it.
Let me simplify it.
So we have our vertical component
is equal to-- now we have a 5 square roots of 2
in the numerator and one in the denominator
and they cancel out.
So we get the square root of 3 times our vertical component,
the magnitude of our vertical component.
It's actually also specifying, well, the magnitude,
we can say, for now, although I'm not using that notation.
So then we have minus 4.9 times this quantity squared.
So that's going to be, the square root of 3 squared
is 3 times the vertical component squared.
And then over 5 squared, which is 25, times 2.
That's the square root of 2 squares.
So 25 times 2 is 50.
And so we get, if we were to simplify this a little bit
more.
Let's subtract this from both sides.
We get-- I'll do it all in one color-- 0
is equal to the square root of 3 minus 1
times our vertical component.
Because if we subtract this from both sides,
that's square root of 3 times our vertical component
minus 1 times our vertical component.
So that's the square root of 3 minus 1 times
our vertical component.
And then we have all of this business.
Minus 4.9 times 3 over 50 times our vertical component squared.
And lucky for us, we can just factor out
one of these s sub y's over here, one of these vectors.
And so we can get-- and I'll just do that in place so that I
don't-- well, let me just-- I don't want to skip many steps.
So this is equal to the square root of 3 minus 1, minus 4.9.
Let me do it in that color.
So let me do it like this.
It's equal to the square root of 3 minus 1 minus 4.9 times
3 over 50 times one of these, times our vertical component.
And then we factored one of those vertical components out.
So we factored one of them out.
So the vertical component of our displacement
could either be-- so we have the product of two things that
equal 0.
So our vertical displacement could
be 0, which is true, because at some point in the path
we literally had 0 vertical displacement.
That was literally where we started.
But that's not the answer we're looking for.
We're looking for this vertical displacement.
So either this is going to be 0.
But that's just kind of the obvious answer.
Or all of this business is going to be equal to 0.
But this is pretty easy to solve for 0 over here.
So we get square root of 3 minus 1 minus 4.9.
Let me just calculate all of these things
just so that I don't have to keep writing them.
So we get the square root of 3 minus 1 is equal to 0.73205.
So I'll just write 0.732 here.
So this is equal to 0.732.
That's that part right over there.
And then 4.94.
I know this problem is getting long.
Just pause it and take a break if you're getting tired.
4.9 times 3 divided by 50 is equal to 0.294.
So minus 0.294.
I could put a 0 out front just to make it clear where we are.
Times this, times our vertical component.
This could also be equal to 0.
Either this is 0 or that is 0.
When this is 0, it gives us the obvious answer.
We're more interested in this.
To solve for this, we can add this to both sides.
And we get 0.732 is equal to negative 0.294
times the vertical component.
We are in the home stretch.
We divide both sides by this, by negative 0.29.
Oh sorry, this will now be positive.
I almost made a careless mistake after 16 minutes of video.
So now we divide both sides by 0.294
and we get our vertical displacement.
So this, I think, a drum roll might be in order.
So we have 0.732 and all of this business,
but I'll just round it there, divided by 0.294 gives us
a vertical displacement of, if we round, 2.50 meters.
Or 2.49 meters, I should say.
So this is equal to 2.49 meters.
This is exciting.
This is equal to 2.49 meters.
And now we can figure out the horizontal displacement
pretty easily, because we know that
the horizontal displacement is square root of 3 times
the vertical displacement.
So let's figure that out.
That's the vertical displacement.
Let's multiply that times the square root of 3.
And we get 4.31 meters.
So we get the horizontal displacement.
The horizontal displacement is equal to 4.31 meters.
So this is equal to 4.31 meters.
So we actually now know the total displacement
in the vertical direction and in the horizontal direction.
And I'll leave it up to you.
If you wanted to figure out exactly how far along the hill
we traveled, you can just use both
of these values in the Pythagorean theorem
to essentially figure out the hypotenuse
of this right triangle.