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Studying for a test? Prepare with these 2 lessons on Two-dimensional motion.
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Let's do another example where we're projecting something and it lands at a different level Let's 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 direction Let's say we're launching something from ground level and we're going to launch it at a pretty step angle over here Let's say we launch it with an angle of 80 degrees and it is going to be going at 30 meters per second So that's the length of the vector; that's the magnitude of that vector And let's say we want to make it land on this landing And this landing over here has a height of 10 meters So what I want to do first of all is figure out how far along the landing do I actually land? And maybe I'll add some other information right here From the launching point to the beginning of the landing-- let's say this right over here is 2 meters So we just want to know how far along the landing do we land So like we did before, we want to break this vector into its horizontal and vertical components I'm gonna go a little bit faster in this video hopefully we're getting hold of this type of thing So the vertical component of our velocity is going to be equal to the magnitude of our total velocity 30 m/s And the horizontal component of our velocity is going to be Again cosine is adjacent over hypotenuse I'm gonna skipping steps. In the last two videos I go into this in a much more detail So 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 displacement If we want to figure out time in the air, we know that displacement is equal to the initial velocity times time-- let me write change in time, that's technically more correct plus acceleration times change in time squared over two 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 much time in the air and then 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 determines this time in the air So we know the acceleration. Remember the convention when we're dealing with the vertical dimension is, up is positive, down is negative So this is -9.8 m/s squared And then what is the total displacement that we're gonna 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 m So this value right here is going to be 10 m Times our change in time So this is -9.8 divided by 2, so it's -4.9 m/s squared, times delta t squared And then we can subtract 10 from both sides and write this into a traditional quadratic equation form So we get -4.9 times delta t squared + 29.54 times delta t -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 -29.54 of 29.54 squared, B squared -4 times A which is -4.9, the negative times and negative is positive, so it's +4 times +4.9 times--I shouldn't have jumped so fast to get rid of the negatives So it's gonna be -4 times A which is -4.9, times C which is -10 So just A times C, -4.9 times -10 These two guys, their signs are gonna cancel out All of that over 2A, over -4.9 times 2, so -9.8 And like we saw in the last video, we want a positive value for this and 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 have a negative value up here And if we already have a negative value here and if we subtract from that negative value we'll definitely have a negative value here Then you divide by a negative value. You'll get a positive value So we can really focus on the subtracting the radical You can try it out. If you try the positive version you'll get a negative value for this entire thing You can try that out after this video just to verify that that will get a nonsensical answer So let's use the negative right over here. So we have -29.54 - square root of 29.54 squared -4 times (-4.9)ĄÁ(-10) which is 49, times 49 Actually I should add some parentheses. Okay Times 49. So this right over here would give me the numerator if I evaluate this We've got a negative value, and I divide that by -9.8 Gives me 5.67 second And you can keep the units in there and make sure the dimensional analysis works You'll find that it does So our total time in the air is 5.67 seconds Now what I want to do, the whole point of this, is to figure out how far along the this landing we land Well, the horizontal component of our velocity is right over here We know that our displacement in the horizontal direction will be our velocity in the horizontal direction-- it's a constant velocity So it's the same thing as our average velocity in the horizontal direction Times the change in time I won't write the unit. This is m/s times s and it'll give us the answer in m Gives us 29.53 m So our total horizontal traveling displacement is 29.53 m It's a vector. That is our horizontal displacement, which is 29.53 m Now we've done a lot of deconstructing vectors What I'm interesting in this video is to construct a vector So we know our horizontal displacement; we also know our vertical displacement It's positive 10 m So what's our total displacement? Let me write this down So we have a horizontal displacement of 29.53 m and we have a vertical displacement of +10 m So what is our total displacement? We can use the Pythagorean theorem here The square of the magnitude of our total displacement is going to be equal to the sum of these two squares This is just the Pythagorean theorem So let me write it over here. So this is the magnitude of our displacement right over here The magnitude of our total displacement squared is going to be equal to 10^2 +29.53^2 To solve for this, we just take the square root of both sides If we just take the square root of both sides we will get the magnitude of our total displacement-- let's get the calculator out to do that So the magnitude of our total displacement is the square root of 10^2 is just 100 + 29-- I can use all this information I'll use the Ans, which literally means the previous answer which is 29.53 squared Gives us a total displacement of 31.18 m Of course it's a vector. This is only the magnitude. We also need the direction So one way to specify 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 functions over here We can use pretty much any of the trig functions But we know the opposite side is 10; we know the hypotenuse here is 31.18 So why not use sine? Sine is opposite over hypotenuse So we know the sin of theta is going to be equal to 10/31.18 Or if you want solve for theta you take the arcsine or the inverse sine of both sides Theta is equal to the inverse sine, or arcsin of 10 / 31.18 Once again get the calculator out to figure out that value I'll take the inverse sine--this is the same thing as arcsin This says, give me the angle. When I take its sine, I get this value So the inverse sine of 10 divided by our previous answer 31.18 is equal to-- This says, give me the angle whose sine is 10/31.18 Here we've constructed a vector. We took its vertical component and its horizontal component and we're able to figure out the total vectors This projectile in this situation, its total placement--just to make it clear Its path will look something like this And we've just calculated its total displacement And I realize that when I started this problem, I asked you I think I was asking, how far along the platform We figured out its total horizontal displacement So if you want to know how far along the platform, the platform starts 2 m to the right so it's really 27.53 m along the platform is where it lands