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Current time:0:00Total duration:15:12

Launching and landing on different elevations

Video transcript

let's do a slightly more complicated two-dimensional projectile motion problem now so in this situation I am going to I am going to launch the projectile off of a platform off of a platform and then it is going to land on another platform it is going to land on another platform and I'm going to fire the projectile I'm going to fire the projectile at an angle let me draw this a little bit better so I'm going to fire the projectile at an angle of let me use a not-so-clean number let's say it is 53 it's a 53 degree angle and it's coming out of the cannon so and the and let me make it clear so it's coming out let me do it this way just to make it 100% clear so so this angle right over here is 53 degrees 53 degrees and we are going to have it come out of the muzzle of the cannon with a velocity of 90 meters per second 90 meters per second and just to give ourselves a sense of the heights or how high it's being launched from so the from the muzzle of the cannon down to here so this height right over here this height right over there let's say that that is 25 meters and let's say that this height right over here this height right over here let's say that that is 9 meters 9 meters and so we're essentially launching this from a height of 25 meters I know in the last video and the last video even though I drew the cannon like this we assumed that it was being launched from an altitude of 0 and then landing back at an altitude of 0 here we're assuming we're launching it from an altitude of 25 meters because that's when it's leaving the muzzle and it's going to start decelerating at least in the vertical direction as soon as it leaves the muzzle and then we're assuming it's not going to land back at the same altitude it's going to land at a different altitude so how do we think about this problem so the first thing you'll always want to do is is divide your velocity vector into its horizontal and vertical components you use the vertical component to figure out how long it's going to stay in the air and then you use the horizontal component to figure out given how long it's in the air how far did it travel and once again we're going to assume that air resistance is negligible so just based on what we did in the last video and I'll work through it relatively I'll go through all of the steps in this one as well the vertical so if we draw our vector we're draw our vector the length here is going to be 90 the angle over here this is our velocity vector the angle over here between the x-axis and our vector is 53 degrees and let me draw the horizontal component the horizontal component would look like this and the vertical component I'll do it in orange the vertical component would look like that's not Orange the vertical component will look like the vertical component would look like this and so the vertical component of the vector what would be the length of this side right over here well this is the opposite side this is the opposite side we know from from basic trigonometry sine of an angle is opposite over the hypotenuse so we know that we know that the sine the sine of 53 degrees is equal to this opposite side is equal to is equal to the vertical velocity is equal to the vertical velocity and it I won't well is equal to the vertical velocity and I could write it it's the magnitude of the vertical velocity I'll write that subscript Y because we're in the Y direction that's the vertical direction over the length of the hypotenuse over the magnitude of our original vector or we can get we can get that this side right over here if we multiply both sides by 90 we get that the magnitude of that side is going to be equal to is equal to 90 times the sine of 53 degrees 90 times the sine of 53 degrees now if we want to if we want to do the horizontal component the horizontal side is adjacent to this cosine sohcahtoa cosine is adjacent over hypotenuse is it J and over hypotenuse so the component the the horizontal component of our velocity I'll say in the X direction over the hypotenuse over 90 over 90 is equal to the cosine of 53 is equal to the cosine of 53 degrees cosine is adjacent over hypotenuse adjacent that's this length over 90 multiply both sides by 90 you get that the horizontal component the horizontal component is equal to 90 is equal to 90 times cosine of 53 degrees times the cosine of 53 degrees now how do we figure out how long this thing stays in the air well we'll use the vertical component for that we'll use the vertical component for that and since we're dealing especially since we're dealing with different levels we can't use that more basic reasoning that hey whatever velocity we start off at it's going to be the same velocity but in the opposite direction or the same magnitude of velocity but the opposite direction because we're not going to the same elevation but we could do is we can use the formula that we derived in the previous in the previous video that the displacement let me just copy and paste this a little bit lower so we can use it so let me copy it and then let me paste it so I'll stick it right over here so we could use this we know that the displacement is equal to the initial velocity and we're dealing with the vertical direction right here times the change in time plus the acceleration times the change in time squared divided by two so how do we use this to figure out how long we're in the air so what is the displacement if we're starting at 25 meters we're starting at 25 meters high and we're going to 9 meters high so over the course while this thing is traveling it will displaced it will be displaced downwards 16 meters it will be displaced downwards 16 meters or another way to think about it is our displacement our displacement in the vertical direction is going to be equal to negative 16 meters let me write that a little bit bigger negative 16 meters right because 25 minus 9 is 16 and so we can put that into the formula that we derived in the previous video we get negative 16 I won't write the unit's here just so that we don't make things too - we don't take up too much real estate so that at least it looks simple is equal to the initial velocity we're dealing with just the vertical dimension here so we're just dealing with the vertical we're just dealing with the vertical here and remember there could it's negative because it's our displacement it's going to be downwards where we're losing altitude so our vertical velocity we already figured that out it is 90 times the sine of 53 degrees 90 times the sine of 53 degrees actually let me do it in that same color the first time we do it you type a problem it's good to know what so 90 times the sine of 53 degrees times our change in time times our change in time is equal to the acceleration of of due to the force of gravity for an object in freefall is going to be negative 9.8 meters per second squared but we're dividing that by 2 so we're going to have negative 4 point 9 minus 4.9 meters per second squared times delta T squared times our change in x squared times our change in x squared so how do we solve something like this there you can't just factor out a t and solve it but you might recognize this is a quadratic equation right over here and the way you solve quadratic equations is you get everything onto one side of this equation and then you either factor it out but more likely in this situation you we will use the quadratic formula which we proved in other videos and can hopefully give you the intuition for it to actually solve for the times where your elevation where your displacement in the vertical direction is negative 60 meters and you'll get two solutions here and one of the solutions will be a negative change in time so it'll be like if you're going in this at some time in the past you are also at negative 16 meters that's nonsensical for this problem so we'll want to take the positive value here so let's put all of this on one side of the let's put it all on one side of the equation so let's add 16 to both sides let's add 16 to both sides on the tan side you just get a zero zero is equal to and I'll write it in kind of the traditional way that we're used to seeing it I'll write the highest degree term first so negative 4.9 times delta T squared times delta T squared and then we have plus ninety sine of 53 degrees plus 90 sine of 53 degrees ninety sine of 53 degrees times delta T times delta T times delta T and then plus 16 plus 16 when do that in yellow plus sixteen all of this is equal to zero and this once again is just a quadratic equation we can stop find its roots then and the roots will be in terms of delta T we can solve for Delta T using the quadratic formula so we get delta T delta T and if this is very unfamiliar to you review the videos on Khan Academy the algebra playlist on the quadratic formula if you don't know where it came from we also prove it for you so it's equal to negative B B is this term right here the coefficient on the delta T so it's going to be negative 90 sine of 53 degrees I'll write the quadratic formula up here for those of you who aren't who don't fully remember it so if I'm trying to solve a x squared plus BX plus C is equal to 0 the roots over here are going to be negative B plus or minus the square root of b squared minus 4ac all of that over all of that over 2 times a these are going to be the X values that satisfy this equation up here so that's all I'm doing over here this is the B value negative B negative B plus or minus plus or minus it's going to turn out that we only care about the plus 1 because that's going to give us the positive value but I'll just write it out here plus or minus the square root the square root of B squared so it's this quantity squared so it is 90 sine of 53 degrees squared minus 4 we're going to run out a little space so minus 4 minus 4 times a which is negative 4.9 so we can make it this well let me just write negative 4.9 times C see over here is 16 times 16 let me put this radical all the way over here all of that over 2a all of that over 2a so a is negative 4 point 9 2 times a is negative negative 9.8 so now we can get the calculator out to figure out our change in time and I'm just going to focus on the positive version of it I'll leave it up to you to find the negative version and see that it'll give you a negative value for change in time and that's nonsensical so we only care about the positive change in time where we get to a change in or we get to a displacement of negative 16 meters let's get the calculator out so we get I'm going to do this carefully we have negative 90 sine of 53 degrees plus I'm doing the plus version here because that will give us a positive value plus the square root of the square root of and I'll do this in parenthesis 90 sine of 53 degrees squared that's that part right there these two negatives cancel out so I could say this is plus 4 times positive 4.9 so plus 4 times 4.9 times 16 and then that closes off our entire radical and so this will give me the numerator up here that gives me the numerator and then I want to divide that by by I want to divide that by did I do the negative 90 oh and I just realized that I made a mistake I I said that the positive version would give you would give you the positive time but now we realize that's wrong because when I took the positive version when I put a plus up here I get a positive 2.1 for for the numerator but then we divide it by negative 9.8 we're going to get a negative value so that's not going to be the time that we care about we care about the time where this is a negative value so let me re-enter that so let me do the negative value so let me move back a little bit and then let me let me replace this with a minus sign replace this with the minus sign so I look at the negative value because I want the positive time and so now my numerator here is a negative value and so this is actually what we care about we care about the numerator is a negative value you divide by negative 9.8 so you divide by negative 9.8 and you get 14.8 I'll just round 14.8 9 seconds so this is delta T the positive version is equal to fourteen point eight nine seconds and so my initial comment about yet wanting to using the positive version was wrong because we have this denominator that's negative so you want the numerator to be negative and only when the numerator is negative well the whole expression be positive and so we got this positive time of fourteen point eight nine seconds I'm going to leave you there and the next part of the video actually I love I might as well just solve instead of making a new video although this is running long so the amount of time that we're in the air the amount of time that we're in the air is fourteen point eight nine seconds so for I were to ask you the horthy the horizontal displacement it's going to be the amount of time we're in the air times your constant horizontal velocity and we already figured out our constant horizontal velocity so if you want to figure out how far how far along the x axis we get displaced we just take that this time times what that just means our previous answer times this value right here times 90 cosine of 53 degrees and that gives 806 meters so this displacement right over here is 806 meters