Learn about Projectile Motion with Ordered Set Notation Video

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Learn about Projectile Motion with Ordered Set Notation

Learn about Projectile Motion with Ordered Set Notation Welcome back and now I want to introduce you really just a different notation for writing vectors and then we’ll do that same problem or slight variation on that problem using the new notation and this is just to expose you the things and so that you don’t get confuse if your teacher uses a different notation than what I’ve been doing. So when we did the unit vectors we learn that we can express a vector as a component of its x and y component so let’s say I had a vector that’s in the last problem I had the vector. So I say I had vector A and that equals to 2(i) + 3(j) that’s the unit vector notation I actually look up on Wikipedia and they actually call that the engineering notation and that’s probably why I use it because I am an engineer or I wasn’t an engineer before managing money but another way to write this and I call this the bracket notation or the ordered pair notation and you could also write it like this. Yeah just one bracket, that’s the x component and that’s the y component it almost like a coordinate pair but since they have the brackets you know it’s a vector but you would draw the exact same way. So given that, let’s do that same problem that we had just done. Hopefully this makes sense to you know. It’s just a different way of writing instead of and I and a J you just write this bracket instead of plus write a comma. So in the old problem, let me draw my coordinate axis again. So y-axis, that’s the x-axis. So in the old problem I started it off with the ball that was 4 feet off the ground. So let’s say that’s four and I hit it at 120 feet per second at a 30 degree angle. So 30 degree angle to the horizontal and there is a fence 350 feet away, that’s 30 feet high. And what we do is figure out what did the ball can clear the fence and we figure out in the last time when we use the unit vector notation that it doesn’t clear the fence but in this problem or the second part of this problem they say that there is 5 meter per second wind gust to the right. So there’s a wind gust of 5 meters per second right when I hit the ball and you could go into the complications of how much does this accelerate the ball or what’s the air resistance of the ball? I think for the simplicity of the problem they are just saying that the x component of the balls velocity right after you hit it increases by 5 meters per second. I think that’s their point. So let’s go back and do the problem the exact same way that we did the last time but we’ll use a different notation. So we can write that equation that I have written before that the position at any given time is a function of t is equal to the initial position. That’s the I right there plus the initial velocity time’s t plus the acceleration vector over 2t². So what’s the initial position and now we’re going to use some of our new notation. The initial position when I hit the ball its x component to zero right. It’s almost like its coordinate and they are not that different of a notation and then the y position is 4, easy enough. What’s its initial velocity? So we can split up the x and y components. The y component is 120sin30degress and then the x component is 120cos30degrees and then they also tell us so that’s just the x component after I hit it but then you say there’s this wind gust that’s going to be plus 5. I think that’s their point when they say that there’s this wind gust. They say right when you hit it for some reason in the x direction and it accelerates a little bit by 5 meters per second. So the velocity vector—this notation actually is better because it takes less base up and you know all this I’s and J’s and pluses confusing everything so the initial velocity vector what’s its x component, its 120cos30. Co30√3/2 times 120 is 60√3 and then you add 5 to it so what is that? Let me just solve it right now. So 3 x √3 x 60 is equal to—and the lets add 5 + 5. So let just round up make it easy since it’s a 109 meters per second 108.9 so let’s just say 109. So the x component of the velocity is a 109 and the y component was just 120 times the sign of 30. Well this is sign of 30 is a half so this is 60. Oh sorry, it should be bracket although some people actually write the parenthesis there so it looks like just like coordinates but I like to keep up with these brackets so that you don’t think that this are coordinates that you know these are vectors and the position vector really is the same thing as the position coordinate but the velocity vectors obviously not a coordinate and then what’s the acceleration vector? Well the acceleration vectors as we said goes straight down and since 32 ft/secs² that’s the acceleration of gravity on earth so the acceleration factor is equal to—it has no x component and its y component is minus 32. So now lets put this back in the original equations. So our position vector, our position is a little arrow in or one side arrows equals my initial position and that’s 04 + my initial velocity vector 109 60 times t + 80² over 2. So t² over 2 x my acceleration vector 0 - 32 and it should be—this is actually a little cleaner way of writing it but this is exactly we did when we do it with unit vectors instead of finding the I’s and J’s we’re just writing the numbers in bracket here. So let’s see if we can simplify this. Okay so our position vector t is equal to—let us spread that (0, 4) + and now we can distribute this t multiply it to x both of this + 109t 60t + and we can distribute this t² over 2 well that times 0 is 0 and then that times minus 32 is -16t² and now we can add the vectors. So the position at any t—so let’s add all the x components of the vectors 0109t is 0. So we just get 109t and then what’s the y component, 4 + 60t - 16t² and there we go and we’ve defined the position vector and the function at any time. So let’s solve the problem. Now that they had this wind gust in our x velocity it’s going a little faster let’s see if we can clear the fence. So how long is it take to get the 350 feet in the x direction? Well this number right here is equal 350. So we have 109t has to be equal to 350 and so what’s 350 divided by 109 is = 3.2 seconds. T is equal to 3.2 seconds and so what’s the height of 3.2 seconds? So lets square that, so let me 3.2 x 3.2 = x 16 = 163. Let’s just say a 164 so this equals 164 and then what 60 x 3.2 is = 192. So what do we get? We get 192 + 4 - 164 so we got 32. So our position vector at time 3.2 seconds is equal to 350 feet in the x direction and 32 feet in the y direction and that will clear that 30 foot fence, our balls going to be 2 feet above the fence. Hope I didn’t confused you too much. See you soon.

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Learn about Projectile Motion with Ordered Set Notation

Welcome back and now I want to introduce you really just a different notation for writing vectors and then we’ll do that same problem or slight variation on that problem using the new notation and this is just to expose you the things and so that you don’t get confuse if your teacher uses a different notation than what I’ve been doing.

So when we did the unit vectors we learn that we can express a vector as a component of its x and y component so let’s say I had a vector that’s in the last problem I had the vector. So I say I had vector A and that equals to 2(i) + 3(j) that’s the unit vector notation I actually look up on Wikipedia and they actually call that the engineering notation and that’s probably why I use it because I am an engineer or I wasn’t an engineer before managing money but another way to write this and I call this the bracket notation or the ordered pair notation and you could also write it like this. Yeah just one bracket, that’s the x component and that’s the y component it almost like a coordinate pair but since they have the brackets you know it’s a vector but you would draw the exact same way. So given that, let’s do that same problem that we had just done. Hopefully this makes sense to you know. It’s just a different way of writing instead of and I and a J you just write this bracket instead of plus write a comma.

So in the old problem, let me draw my coordinate axis again. So y-axis, that’s the x-axis. So in the old problem I started it off with the ball that was 4 feet off the ground. So let’s say that’s four and I hit it at 120 feet per second at a 30 degree angle. So 30 degree angle to the horizontal and there is a fence 350 feet away, that’s 30 feet high. And what we do is figure out what did the ball can clear the fence and we figure out in the last time when we use the unit vector notation that it doesn’t clear the fence but in this problem or the second part of this problem they say that there is 5 meter per second wind gust to the right. So there’s a wind gust of 5 meters per second right when I hit the ball and you could go into the complications of how much does this accelerate the ball or what’s the air resistance of the ball? I think for the simplicity of the problem they are just saying that the x component of the balls velocity right after you hit it increases by 5 meters per second. I think that’s their point.

So let’s go back and do the problem the exact same way that we did the last time but we’ll use a different notation. So we can write that equation that I have written before that the position at any given time is a function of t is equal to the initial position. That’s the I right there plus the initial velocity time’s t plus the acceleration vector over 2t². So what’s the initial position and now we’re going to use some of our new notation. The initial position when I hit the ball its x component to zero right. It’s almost like its coordinate and they are not that different of a notation and then the y position is 4, easy enough. What’s its initial velocity? So we can split up the x and y components. The y component is 120sin30degress and then the x component is 120cos30degrees and then they also tell us so that’s just the x component after I hit it but then you say there’s this wind gust that’s going to be plus 5. I think that’s their point when they say that there’s this wind gust. They say right when you hit it for some reason in the x direction and it accelerates a little bit by 5 meters per second.

So the velocity vector—this notation actually is better because it takes less base up and you know all this I’s and J’s and pluses confusing everything so the initial velocity vector what’s its x component, its 120cos30. Co30√3/2 times 120 is 60√3 and then you add 5 to it so what is that? Let me just solve it right now. So 3 x √3 x 60 is equal to—and the lets add 5 + 5. So let just round up make it easy since it’s a 109 meters per second 108.9 so let’s just say 109.

So the x component of the velocity is a 109 and the y component was just 120 times the sign of 30. Well this is sign of 30 is a half so this is 60. Oh sorry, it should be bracket although some people actually write the parenthesis there so it looks like just like coordinates but I like to keep up with these brackets so that you don’t think that this are coordinates that you know these are vectors and the position vector really is the same thing as the position coordinate but the velocity vectors obviously not a coordinate and then what’s the acceleration vector? Well the acceleration vectors as we said goes straight down and since 32 ft/secs² that’s the acceleration of gravity on earth so the acceleration factor is equal to—it has no x component and its y component is minus 32.

So now lets put this back in the original equations. So our position vector, our position is a little arrow in or one side arrows equals my initial position and that’s 04 + my initial velocity vector 109 60 times t + 80² over 2. So t² over 2 x my acceleration vector 0 - 32 and it should be—this is actually a little cleaner way of writing it but this is exactly we did when we do it with unit vectors instead of finding the I’s and J’s we’re just writing the numbers in bracket here. So let’s see if we can simplify this. Okay so our position vector t is equal to—let us spread that (0, 4) + and now we can distribute this t multiply it to x both of this + 109t 60t + and we can distribute this t² over 2 well that times 0 is 0 and then that times minus 32 is -16t² and now we can add the vectors. So the position at any t—so let’s add all the x components of the vectors 0109t is 0.

So we just get 109t and then what’s the y component, 4 + 60t - 16t² and there we go and we’ve defined the position vector and the function at any time. So let’s solve the problem. Now that they had this wind gust in our x velocity it’s going a little faster let’s see if we can clear the fence. So how long is it take to get the 350 feet in the x direction? Well this number right here is equal 350. So we have 109t has to be equal to 350 and so what’s 350 divided by 109 is = 3.2 seconds. T is equal to 3.2 seconds and so what’s the height of 3.2 seconds? So lets square that, so let me 3.2 x 3.2 = x 16 = 163. Let’s just say a 164 so this equals 164 and then what 60 x 3.2 is = 192. So what do we get? We get 192 + 4 - 164 so we got 32. So our position vector at time 3.2 seconds is equal to 350 feet in the x direction and 32 feet in the y direction and that will clear that 30 foot fence, our balls going to be 2 feet above the fence. Hope I didn’t confused you too much. See you soon.

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