The second part of this question asks us
how high does the subject go as it was flying.
What was the maximum height that it reached?
So there's a few ways to do this problem, just as I sort of pointed out a couple of the ways,
we could have done the range problem, the range part of this problem.
So let me go over, one way to do this,
I may be mention some of the other ways to think about it as we go.
So to find the maximum height, we again, have the exact same sort of situation.
We have an object launched to an angle, in this case, 45 degrees.
And as it goes, it's going to reach some maximum height.
When it's at its maximum height, there's a few things that we can say about this position,
and it's very important to remember this point.
Becasue as I've discussed before, there are some things about a problem
that you will need to infer on your own or deduce about this nature of a problem
that will not be given to you explicitly. And one of those things is this,
the one in object is that its maximum height as we have here,
it is not moving at all in the vertical direction momentarily.
Right at that moment is moving up, and then right after that moment it's moving down.
So right at the pinnacle right at the peak, there is actually no vertical velocity up or down,
because it's right at that moment where it's transitioning.
So the important thing here is that the velocity in the y direction is equal to zero at that point.
The other thing we do indifferently about this part,
in the first part, is we are considering a different initial and final position.
The initial position here is again where the cannonball started up as it was launched.
The final position now is this apex. We're not taking our final position to be the ground anymore.
So our variable will have changed. So be careful when you change,
which frames you're looking at, which snapshots of your object you're looking at.
Because when you change those, all your variables in your problem can changed as well.
So to find the height of this object, what we're going to do is again catalog some of the things that we know,
including that the vertical position is zero. But that the final vertical position is now unknown.
We don't know how height it goes vertically, and that's what we were trying to find out.
The initial velocity in the wide direction is still what we found which is v times the sine of theta,
and we know as I pointed out that the vertical velocity at that moment is zero.
So the final vertical velocity is equal to zero.
And finally we do know that the acceleration of the vertical direction is still minus g.
So, looking at what we know and what we want to find, which in the question mark here.
Things that we know and things that we don't know,
and looking through your equations of motion, you should ask yourself,
which equation should I use to most easily find this final height?
So I recommend you pause and look through them and see if you can deduce for yourself very quickly,
which equation of the ones that we've introduced you should use of the equations of motion.
If you've done this, and looked over your equations of motion,
what you should have found is that one of them again, did not have time in it.
One of our equations did not have a time variable in it, and that equation is perfect for us.
Because again, we do not know how long it took for this object to go from the initial to the final position.
So, we'd like to use that equation and that one as a reminder, is v squared minus the initial velocity squared,
and these are all going to be in the y direction, since we're considering just this one,
is equal to twice the acceleration times the total distance y minus y zero.
Now again, this y is the thing that we're trying to find in y zero, a, v zero y, and vy, are all things that we know.
So, let's plug this in and see if we can solve the problem.
So first, let's go ahead and rearrange, we can divide both sides by 2a,
and then add y zero to both sides. So we have the y is equal to vy squared
minus v zero y squared over twice the acceleration, and then we've added y zero to both sides.
Now, let's simplify this and plug in the things that we know about this problem.
For example, we know that the initial height is zero,
we know that the velocity vertically at that height is zero.
And again, this is not to say that the horizontal velocity is zero, that is not the case.
It's just the vertical velocity up and down at zero, and that's what we're discussing here.
And then we know that a is minus g, so let's do that as well.
So, we have minus v zero in the y direction, which then we saw is v sine of theta.
So this is equal to minus v sine of theta squared, over 2 times minus g.
So we can plug in these variables, cancel the minus signs, so we have v squared,
sine squared of theta over 2g. It's worth pausing right now, I'm looking at this expression,
right in our sine squared theta.
In this way without putting the squared outside.
Well, this is just the fancy way of saying sine of theta as a quantity, being squared.
We just don't like writing sine of theta squared,
because then it looks like maybe I'm squaring the angle before I take the sine,
and this is not what we're doing. So instead of writing this whole big expression
with the parenthesis everytime, we just put the 2 in front of the sine
instead of the theta, to make sure we understand,
we're not squaring the angle before we take the sine.
So, plugging in finally our values we have, 100 squared times a sine of theta which is square root of 2 over 2
as we saw before squared, divided by twice the gravitational acceleration.
And I'm gonna make this approximate, and again use a value of 10.
Solving this, we again have the square root of 2 over 2 squared is equal to 1/2,
so this will be equal to 1 times 10 to the 2, which is writing this in scientific notation square,
times 1/2 over 2 times 10. Now, the reason I wrote this in the scientific notation like this,
is to make very clear what I was doing in the first part of the problem.
Where we use scientific notation to quickly understand what the value of a product would be.
We're gonna go into a lot of this detail for these first problems,
and then you'll be able to use these tools on your own later as we go.
So, be sure to practice these kinds of things. So now we have 10 to the 4th, since we do 2 square,
divided by 10, so these are powers of 10. So, we'll have 10 to the 3rd over here,
then we have 1/2 times 2, which is 1/4. So, the final answer is 0.250 times 10 to the 3rd,
or another words, 250 meters is our maximum height.
What we're seeing here is how to solve an equations of motion problem,
and we have projectile being launched, and we found both the range as we call it,
how far it goes horizontally. We've also found how high it goes at its maximum height
by using the vertical and the horizontal equations of motion together,
and playing them off each other and their strengths and weaknesses.
For the horizontal part, we needed to find the time from the vertical part and plug that in.
For the vertical part on it's own, as in this case, we were able to just solve it all on its own
without referencing the horizontal equations of motion at all.
So this is a good example problem, I recommend working through this problem,
or problems like it on your own, a few times until you're very comfortable with how to use the equations of motion,
in both dimensions, and how do we use those two dimensions together
as you start searching for, whatever it is, that you're being asked in the problem that is unknown.
And that is our summary of the two dimensional motion both vectors and equations of motion, and thanks for watching.