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# Projectile motion with ordered set notation

## Video transcript

welcome back and I want introduce you really just a different notation for writing vectors and then we'll do that same problem or a slight variation on that problem using the new notation and this is just to expose you to things and so that you don't get confused if your teacher uses a different notation than what I've been doing so when we did the unit vectors we learned that we can express a vector as as a component of its x and y component so let's say I had a vector let's say in the last problem I had the vector I had the vector what was a vector let me just pick a random vector just to show you so let's say I have vector a and that equals I don't know two times the unit vector I plus three times the unit vector J that's the unit vector notation I actually looked it 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 was an engineer before managing money but another way to write this and I call this the bracket notation or the ordered pair notation is you could also write it like this you know just one bracket that's the X component and that's the Y component it almost looks like a coordinate pair but since they have the brackets you know it's a vector but you would draw it the exact same way so given that let's do that same problem that we had just done hopefully this makes sense to you you know it's just a different way of writing instead of an eye in a J you just write these brackets instead of a plus you're at a comma okay let's clear this and I'm going to do a slight variation this was actually the second part of that problem my cousin gave these problems to me and they're they're pretty good so I figure to stick with them so in the old problem let me draw my coordinate axes again so y-axis that's the x-axis so in the old problem I started off with a ball that was four 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 it's a 30 degree angle like that it's a 30 degree angle to the horizontal and there's a fence 350 feet away that's 30 feet high so the fence is 30 feet high so it's roughly right in there that's 30 what we need to do is figure out whether the ball can clear the fence and we figured out in the last time when we use the unit vector notation that it doesn't clear the fence but in this in this problem or the second part of the problem they say that there's a 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 that accelerate the ball or what's the air resistance of the ball I think for the simplicity of the problem they're 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 with last time but we'll use a different notation so we can write that equation but I had written before that the position at any given time the position at any given time as a function of T is equal to the initial position that's an eye right there plus the initial velocity these are all vectors the initial velocity times T plus the acceleration vector over 2t squared 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 0 right it's almost it's like it's coordinate and they're not that different of a notation and then the Y position is for easy enough what's its initial velocity let me do it so we can split up into the X and the y components the Y component is 120 sine of 30 degrees and then the X component is 120 cosine of 30 degrees and then they also tell us so that's just just the X component after I hit it but then they say there's this wind gust so it's going to be plus 5 I think that's their point when they say that there's just wind gusts they say that right when you hit it for some reason in the X Direction it accelerates a little bit by 5 meters per second so what was so the velocity vector this notation actually is better because it takes less space up and you don't have all these eyes and J's and pluses confusing everything so the initial velocity vector what's its X component it's 120 cosine of 30 cosine of 30 is square root of 3 over 2 times 120 is 60 square roots of 3 and then you add 5 to it so what is that let me just solve it right now so 3 times the square root square root of 3 times 60 is equal to and then let's add 5 plus 5 so let's just round up make it easy since it's 109 meters per second hundred 8.9 so let's just say 109 so the X component the velocity is 109 and the Y component was just 120 times the sine of 30 well this is sine of 30 s 1/2 so this is 60 oh sorry it should be brackets although some people actually write the parentheses there so it looks just like coordinates but I like to keep it with these brackets so that you don't think that these are coordinates you know these are vectors and the position vector really is the same thing as the position coordinate but a velocity vector is obviously not a coordinate and then what's the acceleration vector well the acceleration vectors we said go straight that's not straight down this is straight down it says it's minus 32 feet per second squared that's the acceleration of gravity on earth so the acceleration vector is equal to it has no X component and it's Y component is minus 32 so now let's put these back in that original equation so our position vector and I'll switch colors to keep things from getting monotonous our position vector these are little arrows are one-sided arrows with equals my initial position and that's 0 4 plus my initial velocity vector 109 6 T times T now I'm running out of space plus a T squared over 2 so T squared over 2 plus T squared over 2 times my acceleration vector zero minus 32 and it should be this is actually a little cleaner way of writing it but this is exactly what we did when we did it with unit vectors instead of writing the i's and J's we're just writing the numbers and brackets here so let's see if we can simplify this so let me let me draw a line let me write in different colors so that you know what I'm doing okay so our position vector T is equal to now let's write that 0 4 plus and now we can distribute this T multiply it times both of these plus 109 T 60 t plus and we could distribute this T squared over 2 well that times 0 is 0 and then that times minus 32 is minus 16 T squared and now we can add the vectors so the position at any T so let's add all the X components of the vector 0 109 T 0 so we just get 109 T and then what's the Y components for plus 60 t minus 16 T squared minus 16 T squared and there we go and we've defined the position vector at a function of any time so let's solve the problem they had this wind gust and our X velocity is going a little faster let's see if we can clear the fence so how long does it take to get to 350 feet in the x-direction well this number right here has to equal 350 so we have 109 T has to be equal to 350 and so what's 250 / 109 350 divided by 1 0 9 is equal to 3.2 seconds T is equal to 3.2 seconds and so what's the height at 3.2 seconds so let's square that so let me 3.2 x 3.2 equals x 16 equals 163 point let's just 164 so this equals 164 and then what 60 times 3.2 60 times three point two is equal to 192 92 so what do we get we get 192 sorry we have 192 plus 4 plus 4 minus 164 right minus 164 is equal to 32 so our position vector at time 3.2 seconds is equal to 350 feet in the x-direction and 32 32 feet in the y-direction and that will clear that 30 that 30 foot fence our ball is going to be 2 feet above the fence I hope I didn't confuse you too much see you soon