00:02
In this video, i'm going to be looking at 2d kinematics, so motion in two dimensions.
00:08
And what our problem is going to be looking at is two girls playing a game of catch.
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A girl number one throws a ball to girl number two.
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The ball starts at a height of h equals 1 .5 meters above the ground.
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And it has an initial velocity v of 17 meters per second.
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And that's add an angle theta of 38 degrees with respect to the horizontal.
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Second girl is going to catch the ball at the same height it was launched from, so that also equals 1 .5 meters.
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That's my final height.
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Let's call this initial height, h -o.
00:49
And what i want to find are the initial and final components of the velocity of this ball in the x and y directions.
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So how do we do this? i know my initial x velocity simply equals the magnitude of my velocity times the cosine of my launch angle.
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And i know my initial y velocity is going to be the magnitude of my velocity times the sign of theta.
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All right.
01:23
So my initial velocity was 17 meters per second.
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My angle is 38 degrees.
01:28
So this equals 13 .4 meters per second, and that's in the x direction.
01:38
And this is 10 .5 meters per second.
01:43
That's in the y direction.
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My initial velocity, the o vector, equals 13 .4x plus 10 .5y.
01:56
And of course that's in meters per second.
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Now i want to find the final components of this ball's velocity.
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So the velocity just before the second girl catches it.
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I know in the x direction i have no acceleration.
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I have no forces added.
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So my velocity is going to be constant.
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So vxf equals the same value as vx initial 13 .4 meters.
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Per second.
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Of course, that's in the x direction.
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In the y direction, i know that i'm launching from the same height that i'm going to be catching the ball at.
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So my change in height will be zero and my vy final is simply the negative of my initial velocity in the y direction.
02:46
So that's just negative 10 .5 meters per second in the y direction.
02:52
So my final velocity vector looks like 13 .4x minus 10 .5y and meters per second.
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Next, i want to find how long the ball is in the air.
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So i'm going to use this equation in the y direction.
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The final equals the initial plus acceleration times time.
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And solving this for t, i get t equals negative 10 .5 meters per second minus my initial velocity minus 10 .5 meters per second over my acceleration.
03:34
That's negative g, so negative 9 .81 meters per second.
03:38
See the negatives drops out.
03:41
This is just on the top.
03:42
We have 10 .5 plus 10 .5.
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So we have 21 divided by 9 .8 per time and that equals 2 .14 seconds.
03:54
Right, so that's the total flight time of the ball in going from the first girl to the second girl.
04:00
Next, we want to find the horizontal distance between these two girls, right, and that's easy.
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We know that dx simply equals v of x times t.
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We know that this v is constant, so we can use this equation.
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So this becomes my x component of the velocity.
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That was 13 .4 meters per second times my 2 .14 .4 .7.
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Seconds times 2 .14 seconds equals a distance d of 28 .7 meters.
04:33
Okay, now for the second part of this problem, the second girl has caught the ball and she's throwing it back to her friend.
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So she's going to throw it back to the first girl, but she throws it a little bit too fast.
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And it's going to pass some distance directly over the first girl's head.
04:51
And we want to find that velocity...