00:01
In this problem, we have two distinct throws.
00:06
One where julie is throwing it to sarah, and then sarah throws it back in the general direction of julie.
00:12
Now, let's go through this diagram in detail.
00:16
So julie here is on the left.
00:18
It's really where julie and sarah are the balls, really what we're tracking, and then we're marking where the ball could be caught.
00:27
So really is the ball's position that really matters.
00:29
So really, let me put a circle around that to indicate that it's service position and the ball being caught.
00:36
All right, let's start from the top here.
00:39
X and y, positive x to the right, positive y upward, acceleration in x, no air resistance, so zero, acceleration in y minus g.
00:49
Now let's look at the motion, the initial position.
00:54
We made that, so x0 is equal to zero, where julie is, where the ball is being released.
00:59
We're told that leaves her hand 1 .5 meters above the ground, and we'll make it so that the time is zero at that point.
01:05
We're told that the initial speed is 17 meters per second and angle 38 degrees to the horizontal.
01:11
So that's julie's portion.
01:14
Now here it's caught by sarah.
01:16
We do not know where sarah is horizontally.
01:19
We do know that it's caught 1 .5 meters above the ground, and we do not know the time it takes to go from the throw, julie's throw, to being caught.
01:28
We'll have to find all those things.
01:30
But that's the diagram.
01:32
And from that, pretty much everything follows.
01:35
If you have the diagram.
01:36
All right.
01:37
Next, we need the components of the initial velocity for julie.
01:44
The initial velocity of the ball.
01:47
V -0 .x is the adjacent side.
01:54
It's going to be v .0.
01:56
Cosine theta.
01:58
Sign is the opposite.
02:01
That is v -0 .y.
02:08
This, if you want to, you could calculate it.
02:10
I would leave it to the end and only calculate when you need to, but it's up to you.
02:15
So we have that.
02:16
Now, we're going to need, we know that there's going to be a throwback.
02:22
So i think it's wise that we know x1.
02:26
But to know x1, we're going to need t1.
02:28
So we've got a few things to do.
02:31
So let's look at what goes on in y.
02:33
The ball starts 1 .5 meters above the ground and is 1 .5 meters above the ground.
02:38
From that, we can get the time.
02:42
So, y1, y, naught, plus v0, y, t1, t1.
02:47
This would normally be t1 minus t0, but t0, i made zero, so we don't just write t1.
02:53
Minus one half g, t1 squared.
02:57
That's our equation.
02:58
Now, let's put in anything that is zero or equal.
03:03
Y1 and y0 are both the same.
03:05
They're 1 .5 meters.
03:09
So when you subtract, that gives you zero.
03:14
So from this we get one -half g, t -1 squared is equal to v -0 -y -t -1.
03:22
So this gives me this t -1 goes away and the power goes away.
03:27
That one solution is t -1 equals zero.
03:30
Well, yeah, that's the launch.
03:31
That's why i made it that way.
03:33
So that's certainly true, but of no interest to us.
03:37
So t1 is equal to 2 v0y over g.
03:43
And putting in our expression for v0y, 2 v0 .5, 2 v .0 sine theta over g.
03:51
Now we can put in our values and get a number.
03:54
2 .17 meters per second, sine 38 degrees over 9 .8 meters per second squared.
04:07
And this works out to be 2 .14 seconds.
04:12
So that's how long it takes from julie's throw to being caught by sarah.
04:19
Now, let's find x1, x1, x ,0, x ,0, xt1, plus 1 1ā2axt1 square.
04:32
Let's put in anything that we know or zero.
04:35
X not, i made zero.
04:36
There is no acceleration.
04:37
So this is just v .0x t1, which is v .0 cosine theta t1.
04:46
And we can put in our values here, 17 meters per second, cosine 38 degrees, and 2 .14 seconds.
04:58
And this works out to be 28 .7 meters.
05:04
So that's located sarah's position, which is, key.
05:10
The time was necessary to get sarah's position, but x1 is key as you're going to see in the next diagram.
05:16
So this part of the problem is done.
05:19
Julie throwing to sarah, this diagram is over with.
05:22
We're going to have a completely new diagram now, new set of axes, new labeling for the zero and the final points, and that will use.
05:31
So this we don't even care about it anymore.
05:33
We got what was key.
05:35
This was key.
05:37
That was key.
05:40
So now, that may even make it so you can't even see that part.
05:45
Sarah throwing the julie.
05:49
All right, let's trace our end.
05:50
Notice, i don't have to, but i'm free to choose my axis any way i want.
05:56
So i'm going to make positive just so.
05:57
Just so i don't have to worry about negative signs to make your life more complicated.
06:01
I make the positive x to the left.
06:03
It's a completely independent from the previous picture.
06:08
So we can do this.
06:09
And positive y is still up.
06:11
So the acceleration due to gravity.
06:13
You know, ay still minus g, still minus.
06:17
All right, let's talk about this.
06:19
Sarah, this is julie's horizontal position.
06:21
She's not going to catch it.
06:22
It's going to be up here.
06:24
So sarah throws the ball at some v -0, which we're looking for.
06:27
That's what this big arrow stands for.
06:29
It's a goal.
06:29
That's our goal.
06:33
At an angle theta, you don't know either one.
06:38
It follows the parabolic path.
06:40
It reaches the highest point, and we're told about this.
06:44
We do not know where it is in x at the highest point.
06:47
Are told where it is in y from the ground, from the ground, which is number zero, eight meters above the ground.
06:53
Vmx is the same as whatever v0x is.
06:56
Vmy, though, zero.
06:59
That defines.
07:01
If v, if v, if v, y was not zero, it's still be moving upward or coming down...