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Current time:0:00Total duration:13:46

Video transcript

let's do another example where we're projecting something and it lands at a different level it will also figure out some other interesting things we'll figure out what the actual velocity vector is when it's landing so both its magnitude and its direction so let's do a situation so let's say we're launching something from ground level and we're going to launch it at a pretty steep angle over here so let's say we launch it with an angle let's say we launch it with an angle of let me just let's say it's 80 degrees we launch it with an eight at an 80 degree angle at let me we launch it at an 80 degree angle 80 degree angle and it is going to be going at 30 meters per second 30 meters per second so that's the length of this vector that's the magnitude of that vector and let's say we want to make it land on this landing and this is landing right over here this landing right over here has a height this landing over here has a height of 10 meters this height over here is 10 meters so what I want to do first of all is I want to figure out how far along the landing how far along the landing do I actually land how far along the landing so and maybe I'll add some other information right here so say from this launching point to the beginning of the landing let's say that this right over here is 2 meters 2 meters so we just want to know how far along the landing do we land so like we did before we want if we want to break this vector into its horizontal and vertical components its horizontal and vertical components and I'm going to go a little bit faster in this video because hopefully we're getting the knack for this type of thing so our velocity our velocity in the vertical or our vertical component of our velocity is going to be equal to the magnitude of our total velocity 30 meters per second times and it's going to be the sine of 80 degrees because sine is opposite over hypotenuse times the sine of 80 degrees and our and we'll just get it out of the way right now the horizontal component of our velocity is going to be 30 meters per second and I'm not writing the unit's here just to save some space times the cosine of a once again cosine is adjacent adjacent adjacent over hypotenuse and if you feel like I'm skipping steps in the last few videos I go into this into much more detail so what how much time do we spend in the air how much time do we spend in the air so once again in the last few videos we saw that we can look at our we can look at displacement if we want to figure out time in the air we know that displacement is equal to the initial velocity initial velocity times time plus acceleration let me write times change in time since that's more technically that's technically more correct plus acceleration times change in time squared over over 2 now in our situation we know what our initial velocity is we're talking about the vertical direction right over here so our initial velocity is going to be this we're trying to figure out how time in the air and the vertical component determines that because at some point when it hits back to the ground it's not going to be traveling anymore so that's what's determining its time in the air so we know that we know the acceleration remember our convention when we're dealing with the vertical dimension is up is positive down is negative so this is negative 9.8 meters per second squared and then we can and then what is the total displacement that we're going to have well we're starting at ground level and we're just talking about the vertical remember that so our total displacement is going to be 10 meters so this value right here is going to be 10 meters so it simplifies to 10 meters I won't write the unit's here as equals 10 is equal to what's 30 sine of 80 degrees so we have 30 times the sine of 80 degrees gives us 29.54 29.54 so this is 29.54 times our change in time times our change in time and so this is negative 9.8 divided by 2 so it's negative I'll do that same green color for point that's not the same green color negative 4.9 m/s squared I'm not the units here times delta T squared times delta T squared and then we could subtract 10 from both sides and write this in a traditional quadratic equation form so we get negative and I'm going to flip the sides to negative 4.9 times delta T squared and then we have plus 29.54 times Delta t minus 10 minus 10 is equal to zero and then we can use the quadratic formula to find the roots of this so the Delta T's that satisfy this quadratic equation are going to be negative B so negative 29.54 plus or minus the square root of 29.54 squared b squared minus 4 minus 4 times a which is negative 4.9 so the negative times a negative is a positive so it's plus 4 times positive 4.9 times I shouldn't have I shouldn't have jumped so fast to get rid of the negative so it's going to be minus 4 times a which is negative 4.9 times C which is negative 10 so just a times C negative 4.9 times negative 10 so let me just write negative 4.9 times negative 10 these two guys their signs are going to cancel out all of that all of that over all of that over 2a all of that over negative 4.9 times 2 so negative 9.8 and like we saw in the last video we want a positive value for this a negative time is nonsensical that's kind of going into the past so we want a positive value and since we have a negative in the denominator we want to have a negative value up here and if we already have a negative value here if we subtract from that negative value we're definitely to have a negative value up here and then you divide by a negative value we'll get a positive value so we can really focus on the subtracting the radical and you could try it out if you try the positive version you're going to get a negative value for this entire thing do that on you could try that out after this video just to verify that you'll get a nonsensical answer so let's use the negative right over here so we have negative 29.54 - the square root of the square root of 29.54 squared and then we have minus four minus four times negative four point nine times negative ten these two when you take the product is positive 49 so x times 49 and I should add a I should add a parenthesis over there so let me insert a parenthesis you can insert parenthesis okay so times 49 so this right here will give me the numerator if I evaluate this it got me a negative value and then I divide that by negative 9.8 so divided by negative 9.8 gives me five point six seven five point six seven seconds this is equal to five point six seven seconds and you could keep the units in there and make sure that all of the dimensional analysis works and I think you'll find that it does so our total our total vertical or the total time in the air is five point six the total time in the air is five point six seven seconds now what I want to do the whole point of this is to figure out is to figure out how far along this landing we land well our horizontal component of our velocity is right over here we know that we know that our displacement in the horizontal direction our displacement in the horizontal direction will be our velocity in the horizontal direction and it's a constant velocity so it's the same thing as our average velocity in the horizontal Direction times the change in time well the change in time is so I'll just write it out times our change in time so this is going to be equal to 30 cosine of 80 degrees 30 cosine of eighty degrees times five point six seven seconds times five point six seven seconds and this I won't write the unit's this is meters per second times seconds it'll give us an answer in meters so once again we have this is our time this is a five point six seven times 30 cosine of eighty degrees gives us 29 point five three meters so our total horizontal travelling displacement I guess we could say is 29 I already forgot the number twenty nine point five three meters twenty nine point five three meters and I alright this is a vector that is our horizontal displacement it's equal to twenty nine point five three meters now we've done a lot of deconstructing vectors what I thought would be interesting in this video is to construct a vector so we know our horizontal displacement we also know our vertical displacement it's positive ten meters so what's our total displacement let me write this down so we have a horizontal displacement we have a horizontal displacement of twenty nine point five three meters and we have a vertical displacement of +10 meters we have a vertical displacement of positive ten meters and so what is our total displacement what is our total displacement going to be well we could use the Pythagorean theorem now here the square of the magnitude of our total displacement is going to be equal to the sum of these two squares or another way this is just the Pythagorean theorem so we could have if we call this length over here C or let me just write it this is a magnitude of our displacement right over here the magnitude of our total displacement squared is going to be equal to ten squared plus twenty nine point five three squared so it's going to be ten squared plus twenty nine point five three squared I'm going to do that same color plus twenty nine point five three squared if we want to solve for this we just take this the square root of both sides so I'll just do that in place so if we just take the square root of both sides we will get the magnitude of our total displacement and I can get the calculator out once again to do that so the magnitude of our total displacement is the square root of 10 squared is just 100 plus 29 I can even use well I could use all of this information so I don't even lose some precision I could say second answer that literally means the previous answer which is that twenty nine point five three squared gives us a total displacement of 31 point one eight meters so this is equal to 31 point 1 8 meters and of course it's a vector so we did we've only given you the magnitude we also need the direction so one way to specify a direction is to give you the angle with the horizontal and let's call that angle theta and once again we can use our trig our trig functions over here we could use pretty much any of the trig functions but we know the opposite side is 10 we know the we know the hypotenuse here is 30 1.18 so why not use sine sine is opposite over hypotenuse so we know that the sine of theta is going to be equal to 10 over 30 1.18 10 over 30 1.18 or if you want to solve for theta you take the arc sine of both sides or the inverse sine of both sides you get theta is equal to the inverse sine or I could write arc sine over here of of 10 of 10 over 30 1.8 10 over 30 1.18 I should say so let's once again get the calculator out to figure out that value so I'm going to take the inverse sine once against this is the same thing as the arc sine this is says give me the angle that when I take its sine I get this value right over here so the inverse sine of 10 divided by our previous answer 30 1.18 I'll just say our previous answer is equal to so this says give me the angle whose sine is 10 over 30 1.18 so I have eighteen point seven degrees or 18 point seven one degrees so this is equal to eighteen point seven one degrees eighteen point seven one degrees above the horizontal so here we've constructed a vector we took its vertical component and its horizontal component and we were able to figure out the total vector so this this projectile in this situation is going it's total displacement so just to make it clear its path will look something like this its path will look something like this its path is going to look something like this and we've just calculated its total displacement its total displacement is 30 one point one eight meters eighteen point seven one degrees above the horizontal now the other thing I realized that when I started this problem I asked you I think I was asking you how far along the platform and we figured out its total horizontal displacement so if you want to know how far along the platform the platform starts two meters to the right so it's really twenty seven point five three meters along the platform is where it lands