In the last video, where we launched something at an angle,
we figured out the horizontal and the vertical components
of that launched velocity.
And we used a vertical component to figure out
how long that thing is in the air.
And then we used that time to figure out
how far it will travel, given a constant horizontal velocity.
What I want to do in this video is
to redo part of that problem.
And I want to redo the part to figure out
how long this object stays in the air.
And I'm going to do it in a slightly more complicated way
than the way I did in the last video,
but it's kind of a more powerful way.
And we'll see that it's more powerful in future videos.
So the last video, to figure out how long it's in the air,
we said, look.
We're starting and stopping at the same elevation.
Whatever our vertical velocity is here upward,
we're going to have the same magnitude of velocity,
but it's going to be downward.
We used that insight to figure out the change in velocity,
and then we used that change in velocity
and the constant acceleration due to the force of gravity
on an object in free fall, to figure out the time in the air.
And it was pretty simple mathematically.
What we're going to do now is, to derive a formula that's
useful in physics, although, I like
to always derive it from scratch.
And that's how we're going to derive it over here.
And you've see me derive this formula before.
But just to start off, we need to remind ourselves
that the vertical velocity that we calculated in the last video
was five meters per second.
And given that vertical velocity,
let's figure out how long this thing stays in the air.
So our vertical velocity, our velocity in the y direction,
is five meters per second.
And also as a reminder, our convention,
when we're dealing with vectors in one dimension,
and this is essentially in one dimension
now, since we know it's in the y direction,
is positive means up, negative means down,
when we're dealing with the vertical direction.
When we're dealing with the horizontal direction,
positive means to the right, negative means to the left.
So just to make sure we have those conventions down.
So how can we figure out how long
this thing stays in the air?
So we can remind ourselves that the displacement,
and everything we're talking about now,
is, well, this will actually apply
whether in any one dimension.
But the displacement is equal to the average velocity
times our change in time.
Times the amount of time that elapsed.
And what is our average velocity?
Our average velocity is-- if we assume constant acceleration--
it is our final velocity plus our initial velocity.
Actually let me write it this way,
it'll make it a little bit cleaner.
It is our initial velocity plus our final velocity divided
by 2.
So literally the arithmetic mean of our initial
and final velocities.
And we're going to multiply that times change in time.
Now, what is our final velocity going to be?
Our final velocity-- I'll write it over here.
Our final velocity is the same thing as your initial velocity,
plus acceleration, your acceleration
times how much time has passed by.
This right here is your final velocity.
So let's substitute that back into this, right over here,
and what do we get?
We get our initial velocity, so our average velocity
is now our initial velocity plus our final velocity.
The final velocity is this over here.
So it's our initial velocity plus the acceleration
times the amount of time that passes by.
All of that divided by-- let me do
that same color-- all of that divided by 2.
And then all of this is multiplied
by the amount of time that passes by.
Once again, this is our total displacement.
So this part right over here-- let me do it in magenta--
we have to vi's.
All right.
So that simplifies to two vi's.
2 times our initial velocities.
And so this expression right here simplifies to two times vi
divided by 2 is our initial velocity.
And then we have-- well, I'll just do it all here.
And then I'll multiply it times the time elapse after that.
Plus the acceleration times time,
the acceleration times the amount of time
that's passed by, divided by 2.
And all of this is multiplied by the amount of time
that passes by.
And this is our total displacement.
And we're almost done.
And if you have to take a physics exam under time
pressure, it's not ridiculous to not
have to rederive it every time.
But I want to emphasize it.
You should realize that these formulas come from almost just
common, basic sense principles, and you
should be able to derive them.
But it's good to memorize them if you have a time pressure
exam.
So now if we multiply this delta t,
you get the total displacement is
equal to your initial velocity times your change in time,
plus-- I'll do the same color-- plus your acceleration
over 2 times-- you have a delta t times a delta t.
Remember, delta just means change in time.
This triangle just means change in time.
So change in time times change in time
is change in time squared.
And this expression right here, this is in some physics books
or in some classes, they'll just give you this formula.
But I wanted to just show you that it's
derived from very, very simple principles.
From just, displacement is average velocity times time.
What is average velocity?
What is final velocity?
It's just initial velocity plus acceleration times time.
And you'll get this expression right over here.
So I did this to figure out how much time we're
going to spend in the air.
So how do we do that for this particular problem?
Well, we know our initial velocity,
that's the launch of the vertical component,
because we're dealing with just the vertical component here.
We know our initial velocity.
We know the acceleration due to gravity.
And in this situation, we want to figure out
when the displacement is going to be--
and this is going to be a little bit unintuitive,
but hopefully it makes sense-- when
the displacement is going to be 0.
Because there's going to be two times when the displacement is
going to be 0.
There's going to be the time right when we launch it,
right at time 0 the displacement is going to be 0.
Then this thing's going to be launched and then come back
to the ground.
And then when it comes back to the ground,
the displacement is going to be at 0 again.
So if we set the displacement to be equal to 0,
we should get two answers to that, and one of them
should have a non-zero time component.
One of them will probably be 0, and the other one
will have a non-zero time, and that's
the one that we should go for.
Because that's the second time, that's
the time when we've hit the ground.
So let's apply that.
So we have the displacement is going
to be 0, so 0 is equal to-- our initial velocity is
5 meters per second upwards.
I won't write the units just for the sake of space.
5 meters per second upwards times change in time, plus,
what's our acceleration?
Well, our convention is if somethings going downwards,
if the one dimensional vector is going downwards, it's negative.
So the acceleration is negative 9.8 meters per second squared.
You divide that by 2.
So we get-- actually just let me write it all out.
Actually let me write it like this.
It's negative 9.8 meters per second squared.
I'm not writing the units here just for simplicity,
just so that we don't take up too much space.
And then all of this times delta t squared.
And lucky for us, this one is pretty simple,
we don't have to resort to factoring, or we will factor,
but we don't have to resort to the quadratic formula
or anything like that to solve this quadratic equation, right
over here.
We can factor out a delta t.
So we get 0 is equal to delta t times 5 minus 9.8 over 4.2
is 4.9.
And if you factor out a delta t out of delta t squared,
you're left with a delta t.
So there's two solutions here, There's
two delta t's that will give us 0.
Either this delta t is equal to 0,
so either delta t is equal to 0--
that's just our starting point.
So that's not the time that we're interested in,
where no time has passed by.
Or all of this business could be equal to 0.
If we have two things when we take their product
that equal 0, that means either one or both of them
are equal to 0.
So we have 5 minus 4.9 delta t is equal to 0.
We could add 4.9 delta t to both sides of this equation.
And we will get 5 is equal to 4.9 delta t.
And then we divide both sides by 4.9 to get the elapsed time.
So delta t is equal to, 5 divided by 4.9
is-- we're at the home stretch-- 5 divided by 4.9
is equal to 1.02 seconds.
So it's equal to 1.02 seconds.
And so once again, we got the same exact answer
we got in the last one, we just used a different method.
And that's one of the really fun things about physics
is as long as you're doing logically
consistent correct things, there's
many ways to often get that answer,
but it should be the same answer.
If it's not, you probably did something
wrong in one of those two ways of doing it.