If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:18:17

Video transcript

what I want to do in this video is tackle a problem that would be considered pretty difficult for most first-year physics students and you frankly probably wouldn't be expected to solve a problem like this in most first-year physics class or if you're in an advanced or honors class you might be expected or it might be a bonus problem but it's an interesting type of problem because what we're going to do is we're going to have we're going to launch a projectile on an incline so maybe we're on the side of a hill so that's so it's a hill let me do it in green so let's say we're on the side of a hill like this and let's say that we know the inclination of the hill the hills inclination is 30 degrees is 30 degrees off the horizontal so this is the horizontal right over here so that is the inclination of the hill and then we are going to launch a projectile we are going to launch a projectile at 10 meters per second we're going to launch it at 10 meters per second and the angle with the hill is 15 degrees so at a 15 degree angle with the hill and the reason why this is more difficult than the traditional projectile motion problems is well we can think about it the projectile is going to be launched and it is going to eventually land at some point on the hill but we can't do this the simple figure out how long it's in the air using it's vertical velocity because we don't know what the vertical displacement for this thing is going to be unless we know how far down how far down the hill it lands because the further the further down the hill and lands the higher the vertical displacement so we have to think about both the the horizontal and the vertical displacement at the same time and we'll as we walk through this you'll see how that can be done so in the so I guess the first thing that we really would always want to do whenever you want to try to find solve this type of problem is break up our velocity into both the horizontal and vertical components so the vertical component of our velocity the vertical component of our velocity is going to be the magnitude of a lot of our velocity of our total velocity 10 meters per second times and be very careful here not the sine of 15 degrees but the sine of the angle with the horizontal so times the sine of 45 degrees times the sine of 45 degrees and I go into a lot more detail in previous videos for the sake of time I won't go into it but it really just comes from sohcahtoa if we if we were to draw the vertical component it would look like this it would look like this this is the angle the sine of 45 degrees is equal to the opposite over the hypotenuse or the hypotenuse times the sine of 45 degrees is equal to the vertical component that's where it's coming from let me get rid of some of the stuff I just drew just so it makes it a little bit cleaner and the horizontal component the horizontal component of our velocity is going to be by the same logic 10 cosine of 45 degrees now let's think about what the horizontal displacement is going to be the horizontal displacement and I'm just going to go straight to the formula that we derived in the last few videos the horizontal displacement is going to be our initial [ __ ] sorry let's do the vertical displacement the vertical displacement I could have done the horizontal displacement first but the vertical displacement is going to be our initial vertical velocity which we know as 10 sine of 45 degrees and by the way let's we could just solve that right now what is the sine of 45 degrees sine of 45 degrees is square root of 2 over 2 cosine of 45 degrees is also square root of 2 over 2 so both of these values 10 times square root of 2 over 2 is 5 square roots of 2 so this whole thing right over here is 5 square roots of 2 meters per second meters per second that's our vertical velocity and our horizontal velocity is also our horizontal velocity is also 5 square roots of 2 meters per second so that simplifies things a little bit but anyway we were talking about our vertical displacement our vertical displacement is going to be our initial vertical velocity 5 square roots of 2 times our change in time times our change in time plus plus the acceleration and plus the acceleration well we know the acceleration is it's negative 9.8 m/s^2 something right minus 9.8 I'm not going to write the unit's here so that we save space x times our change in x squared times our change in x squared all of that over 2 we derive this in several videos especially the last view where we do these two dimensional projectile motions so this gives us our displacement in the Y direction and I can simplify this a little bit our displacement in the vertical direction is equal to is equal to 5 square roots of 2 times delta T times let me do it that same times delta T change in time minus 4.9 minus 4.9 times our change in x squared times our change in x squared so we know we have this constraint right over here and now so this gives us our this gives us our vertical displacement as a function of time let's think about our horizontal displacement as a function of time our horizontal displacement is going to be equal to our horizontal velocity which is 5 square roots of 2 times our change in time times our change in time now what can we do next well we have to have some relationship between our horizontal displacement and our vertical displacement and that relationship is going to be given to us by this incline so wherever we land let's say this is where we eventually do end up landing what is going to be our horizontal and well let's think about our horizontal and our vertical displacements and what their relationship would have to be so this is where we land then this would be our we doing the same colors this right here would be that right there would be my vertical displacement I would move that far up and then our and our horizontal displacement will be this right over here will be that length right over there so that is our horizontal displacement so what is the relationship between our vertical displacement and our horizontal displacement and we know that this angle right here is 30 degrees so we can use some basic trigonometry we have a right triangle we know the opposite side from the angle we know the adjacent side and the trig function that uses the opposite and the adjacent is the tangent function so we get the tangent the tangent of 30 degrees is going to be equal to is going to be equal to the magnitude the magnitude of our vertical displacement over the magnitude of our horizontal displacement over the magnitude of our horizontal displacement and the tangent of 30 degrees that's the same thing as the sine of 30 over let me just do it over here so the tangent of 30 degrees is the same thing as the sine of 30 degrees over the cosine of 30 degrees let me do this a little bit neater and the sine of 30 degrees is 1/2 this is equal to 1/2 and the cosine of 30 degrees is the square root of 3 over 2 so this is equal to 1/2 times 2 over the square root of 3 which is equal to 1 over the square root of 3 so we get the magnitude of our vertical component over the magnitude of our horizontal component this is our horizontal component right over here over our horizontal component is equal to 1 over the square root of 3 is equal to 1 over square root of 3 what's useful about this is it gives us a relationship between our horizontal and vertical component or between our vertical and our horizontal components and we can use this constraint right over here to then to then solve for one of these two and let me show you how we'll do it so let's just explicitly write this so if we if we multiply well let's do it this way let us cross multiply here which is really the same thing as multiplying both sides by the square root of 3 and the magnitude of our horizontal component we get we get the square root of 3 times the magnitude times the magnitude and both of these are going to be positive and we know where well let me just write it this way times the magnitude of our vertical component is going to be equal to the magnitude of our horizontal component the magnitude of a horizontal component so right just just like that so we now have a relationship between the length of these two vectors and we can use this relationship to substitute back into the constraints that we already have so this comes the second constraint right over here let's use this information the second constraint says that our horizontal component is going to of our displacement is equal to 5 square roots of 2 times our change in time or another way of thinking about it if we divide both sides by 5 square roots of 2 we get our change in time is equal to our change in time is equal to the horizontal component the horizontal component of our displacement divided by 5 square roots of 2 but we also know that the horizontal component of our displacement is the square root of 3 times the vertical component of our displacement and I'm going to get a little bit here I explicitly wrote the magnitude notation when we start dealing with the either just the vertical or the horizontal component we can I can just write it like this because it's either going to be a positive or negative value and that specifies both the magnitude and the direction so what I'm going to do right over here and obviously the way I've drawn it right over here both of these are going to be positive values it's upwards displacement in the vertical direction so that's positive by our convention and we're moving to the right so that's positive also by our convention so this so I can rewrite I can rewrite this over here as being equal to square root of 3 square root of 3 square root of 3 times our vertical displacement times our vertical displacement and all that's over five square roots of two all of that is over five square roots of two now the whole reason why I did this is where we this expression right here can contains this information contains the ratio between our vertical displacement and our horizontal displacement and it also contains the information of how does the horizontal displacement how does that change as a function of time so this our time is needs to be equal to this so this is our time as a function of our vertical displacement now time is a function of our horizontal displacement and what we can do is we can use this constraint with our original with our original this vertical displacement as a function of time to then solve for our vertical displacement so let's do that let's substitute this business right here for delta T in this top in this top equation right over here so if we do that and I'll write it big we got our vertical displacement we get our vertical displacement right over there is equal to v square root of two times delta T so it's five square roots of two delta T is all of this business over here so five square roots of two times delta T delta T is square root of three times our vertical component all of that really the magnitude of our vertical component over five square roots of two so that's that right there and then oh actually we could look at this one right over here we can use this constraint this is just simplified so that we have minus 4.9 minus 4.9 times delta T squared so delta T squared is this quantity squared I'll just write it out so I don't want to skip too many steps so delta T once again is the square root of three times the vertical component all of that over five square roots of two squared and now what does this give us so now we literally have a quadratic equation with only one variable so we can solve for this but let me rewrite it let me simplify it so we have our vertical component is equal to now we have a five square root of two times five squared in the numerator and one of the denominator they cancel out so we get the square root of 3 square root of three times our vertical component the magnitude of our vertical component it's actually also specifying well the magnitude we could say for now although I'm not using that notation and then we have this hole it's and then we have minus 4.9 times this quantity squared so that's going to be square root of three squared is three times the vertical component squared the vertical component squared and then fought over five squared which is 25 times two that's the square root of two squared so 25 times 2 is 50 and so we get if we were to simplify this a little bit more let's subtract this from both sides we get I'll do it all in one color zero is equal to the square root of three minus one minus one times our vertical component times our vertical component because if we subtract this from both sides that's square root of three times our vertical component minus one times our vertical component so it's the square root of three minus one times our vertical component and then we have all of this business minus 4.9 times three over 50 over 50 times our vertical component times our vertical component squared and lucky for us we can just factor out one of these S sub Y's over here one of these vectors and so we can get and I'll just do that in place so that I don't well let me just I don't want to skip too many steps so this is equal to the square root of three minus one minus 4.9 let me do it in that color so let me do it like this it's equal to it's equal to the square root of three minus one minus 4.9 times three over 50 times one of these times our vertical component and then we factored one of those vertical components out so we factored one of them out so the vertical component of our displacement could either be so that we have the product of two things they equal zero so our vertical displacement could be zero which is true because at some point in the path we literally had zero vertical to play displacement that was literally where we started but that's not the answer we're looking for we're looking for this vertical displacement so either this is going to be zero but that's just kind of the obvious answer or all of this business is going to be equal to zero but this is pretty easy to solve for zero over here so we get we get square root of three minus one minus 4.9 let me just let me just calculate all of these things just so I don't have to keep writing them so we get the square root of 3 square root of three minus one is equal to 0.73 two zero five so I'll just write 0.732 here so this is equal to 0.73 - that's that part right over there and then four point nine four point nine out of this problems getting along but this is a well let's pause it and take a break if you're getting tired four point nine times 3 divided by 50 divided by 50 is equal to 0.2 9 4 so minus 0.2 9 4 minus is 0.294 I could put a zero out front just to make clear where we are times this times our vertical component this could also be equal to zero either this is 0 or that that is your when this is zero it gives us the obvious answer we're more interested in this to solve for this we can we can add this to both sides and we get zero point seven three two is equal to negative zero point two nine four times the vertical component we are in the homestretch we divide both sides by by this by negative zero point two nine by oh sorry this will now be positive almost made a careless mistake after 16 minutes of video so now we divide both sides by zero point two nine for both sides by zero point two nine four and we get our vertical displacement so this I think a drumroll might be in order so we have 0.732 and all of this business but I'll just round it there divided by 0.294 gives us a vertical displacement of two point if we round to point five zero meters or two point four nine meters I should say so this is equal to this is equal to two point four nine meters this is exciting this is equal to two point four nine meters and now we can figure out the horizontal displacement pretty easily because we know that the horizontal displacement is square root of three times the vertical displacement so let's figure that out that's the vertical displacement let's multiply that times the square root of three and we get four point three one meters so we get the horizontal displacement the horizontal displacement is equal to four point three one meters so this is equal to four point three one meters and so if you wanted to figure out so we actually now know the total displacement in the vertical direction and in the horizontal direction and I'll leave it up to you if you wanted to figure out exactly how far along the hill we traveled you can just use both of these values in the Pythagorean theorem to essentially figure out the hypotenuse of this right triangle