There are many problems I have selected, as always a few, in
this case, 7 that I think will help you the most.
You ready?
I am.
There's an astronaut--
problem 3.2--
and the astronaut is on the moon.
And the astronaut is throwing up an object at a certain
angle theta, and the object reaches the height h.
It reaches the highest point at a time t of h.
That is my shorthand notation.
This is the x direction.
And let this be the y direction.
There is an acceleration, g, and I don't even
know what it is.
I didn't even look it up.
We don't need it.
But what is important that whenever you have an
acceleration in the y direction, you better write
down for it minus g.
The maximum distance that it travels in the x direction,
that's called its x max.
And the time that it hits here as we have seen last time is 2
t of h because of the complete symmetry of the parabola.
There is a velocity with which the astronaut
throws up the stone.
And as I said, she does it at a certain angle theta and the
velocity in x direction, which never changes, equals v0
cosine theta.
v doesn't change because there's no acceleration in
this direction.
The velocity in the y direction, which starts off at
v0 sine theta does change.
It comes to a halt here, 0, and then it increases and it
reverses direction.
Now the idea is that this astronaut is going to throw up
this stone with the same angle theta, but with twice the
velocity v0.
And all you're being asked, does the flight last maybe
twice as long or maybe four times as long?
Is the height that it reaches twice as high or maybe four
times as high?
And does it travel maybe four times as far or
maybe twice as far?
And those are issues that we will discuss now.
I will separate x and y as I have always done.
And then you will see that this problem is
not all that tough.
In the x direction there's no acceleration, so the velocity
remains constant.
So v of x is any moment of time t equals
v0 cosine of theta.
That is my equation number 1 and a of x equals 0.
The position in the x direction as any function any
moment in time equals x0, which I will
conveniently choose 0.
So I choose this origin as 0 both for y and for x.
Plus v0 in the x direction times t.
But the v0 at times t equals 0 is the v at any
time in the x direction.
So that's equation number 2.
Now in the y direction, I have that y as a function of time
would be y0, which I also choose 0.
Plus v0y times t minus 1/2 g t squared.
That's equation number 3.
And the velocity in the y direction as a function of
time equals the velocity in the y direction at time t
equals 0 minus g t.
And that is my equation number 4.
So let's first ask the question, when is y equal h?
In other words, when is v of y equals 0?
So I go to equation number 4.
I substitute in here v y equals 0 and I find
immediately that the time that h is reached, which we have
called t of h, equals v0 times the sine of
theta divided by g.
That's the result that I will frame in red
if you can see colors.
The time that it takes to reach the highest point.
So it should be immediately obvious since the time of
flight, the whole trajectory, is 2 t of h is therefore 2
times v0 times the sine of theta divided by g.
That you can now immediately answer the question that if v0
doubles, but if theta does not change that the flights last
twice as long.
It's as simple as that.
Because the duration of the flight is proportional to v0.
Now how high does the object go if we double v0?
Well, I substitute in equation 3, I substitute
t equals t of h.
Because that's the time that it reaches the highest point.
And when I do that--
and you can do that for yourself--
I find that the height that it reaches equals 1/2 v0 squared
times the sine squared of theta divided by g.
And that is proportional to v0 squared.
Let's also put this in a nice red frame.
And so you see here that when you double v0, all things
being equal and g on the moon is the same all the time, so
you don't have to worry about g, notice that when you double
v0, that it goes four times higher.
Then there is one question left.
And that's the one I would like you to deal with.
And that is the distance traveled.
That is the horizontal distance traveled.
And I think you should do that.
So when we double v0, how much further does it land in the x
direction when v0 is doubled?
OK, I don't think this one's too tough, and so let's go on
to our next problem.