Learn about Projectile motion - part 4 Video

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Learn about Projectile motion - part 4

Learn about Projectile motion - part 4 We will now use that equation we just derived to. To go back and solve or at least address that same problem we’re doing before. So, let’s try that equation down again. So let’s write that equation down again. So actually let’s write the problem now. Let say I have the cliff again and so my initial distance zero but it goes down 500 meters. I'm not going to re-draw the cliff because it takes a lot of space upon my limited chalkboard. So we know that the change in distance is equal to -500meters. And I’m still going to use the example where I do not just drop the ball or the penny or whatever I'm throwing off the cliff. I actually throw it straight up so it’s going to go up and slow down from gravity and then go to zero velocity and start accelerating downwards but you can even say decelerating in the other direction. But the initial velocity, vi=30m/s then of course we know that the acceleration is equal to -10m/s2 because acceleration gravity is always pulling downwards and towards the center of our planet. If we want to figure out the final velocity, we could’ve just used the formula in the last video, vf2= vi2+2ad. But now, want I want to do is I’m going to use the formula we’ve learned in the very last video to figure out how long does it take to get to the bottom off― to hit the ground? So, let’s use that formula. We derived to that the change in distance is equal to the initial velocity times time plus acceleration times squared over two. That’s initial velocity. So the change in distance is -500 is equal to the initial velocity. Well, that’s positive going upwards 30-meter per seconds, 30t― I’m not going to write the units right now because it just makes things― I'll ran out of space but you can re-do with the units and see that units do work out. When square time, you just have to square time units etcetera although we’re solving for time, plus acceleration. Acceleration is -10. We’re going to divide it by two, right? So it’s -5t2 so if -500 is equal to 30t+ (-5), we could have said -5t2 and get rid of this plus. So at first you’ll say “There's the t to the first, t to second. How do I solve this?” And hopefully, you’ve taken Algebra 2 or Algebra 1 in some place and you remember how to solve this otherwise you're about to learn the quadratic equation although I recommend you go back and you learn about factoring in the quadratic equation which there are videos on that I've put on YouTube so I hope you watch those first if you don’t remember. But what we can do, let’s put these two right terms on the left-hand side and then we use the quadratic equation in solving. It sounded it’s easy to factor. So, we get the 5t2-30t-500=0. I just took these terms and put them on the left side. And we could divide both sides by five just to simplify things so we get t2-60-100=0. All right, I could do that because 0 divide it by five is just five so it just cleaned it up a little bit. So let’s use a quadratic equation for those of us who need a refresher I'll write it down. So, the roots of any quadratic― in this case it’s t― we’re solving for― t will equal -b±√b2-4ac over 2a where a is the coefficient on this term, b is the coefficient on this term, -6 and c is the constant so it is -100. So, let’s just solve. So, we get t=-b so, negative― this term. Well, this tem is -6. So if we make it negative, it becomes +6 so it become six plus or minus the square root of b squared, so it’s minus six squared, 36, minus four times a. The coefficient on a here and that’s just one times one for ac, c is a constant term minus 100, -4×1×-100×-100 all of that over 2a. A is one again so all of that is over 2.] That is equals 6 plus or minus the square root—this is minus four times minus 100, so this becomes plus so this becomes 36+400 so 6 ±436/2 and this is not a clean number. And if you type it into a calculator, ― something on the order of about 20.9, we can just say approximately 21, and you might want to get the exact number if you’re actually doing this on a test or trying to sense send something to mars. But for all purposes, I think you get at the point. So I’ll say it approximately now because we’re going to be a little off but just to have clean numbers. This is approximately 21, it is a little 20.9. We’ll say 6±― let me just write 20.9/2 so t approximately six. So, let me ask you a question, if I do 6-20.9, what do I get? I get a negative number. And does a negative time make sense? No, it does not. That means it’s somehow in the past― I don’t want to get a little philosophical but the negative time in this context will not make sense So really we can just stick to the plus, Right? Because 6-20 is negative so 6 will just― there’s only one time that solves this in a meaningful way. So time is approximately equal to 6+20.9 so that’s 26.9/2 and that equals what? 13.45 seconds. Time is equal to―. That’s interesting. I think if you remember way back four or five videos ago when we first did this problem, we just dropped the penny straight from the height. In that problem, I gave you the time. I set it to 10 seconds to hit the ground and we worked backwards to figure out that the cliff was 500-meters high. Now, if you’re at the top of a 500-meter cliff or building and you dropped something that has no air resistance like a penny, it has a very little air resistance, it would take 10 seconds to reach the ground assuming all of our assumptions about gravity and whatever. But if you were to throw the penny straight up off the edge of the cliff at 30-meters per second right here, it’s going to take 13.5, roughly 13 and a half seconds to reach the ground. So, it takes a little bit longer and that should make sense because― you should let me― I have time to draw a picture. In the first case, I just took the penny and its motion just went straight down. In the second case, I took the penny. It first went up and then it went down, right? So, it had all the time where it went up and then it kind of went down on a longer distance so it makes sense. This time it is 10-seconds while this time it was 13.45 seconds. So, you can kind of say that it took― well, you actually can't say that. I don’t want to get too involved. Hopefully, this makes sense to you. If you got a smaller number here, you would have gone and checked your work because why would it take less time when I throw the object straight up. So hopefully, that gave you a little bit of more intuition and you really do have a lot in your arsenal now, all of the equations. And hopefully, the intuition you need to solve basic projectile problems. I'll now do probably a couple of videos where I just took a bunch of problems. It just really, really drives the point home and then I'll expand these problems to two dimensions in angles. And before we get there you might want to refresh your trigonometry. I'll see you soon.

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Learn about Projectile motion - part 4

We will now use that equation we just derived to. To go back and solve or at least address that same problem we’re doing before. So, let’s try that equation down again. So let’s write that equation down again. So actually let’s write the problem now. Let say I have the cliff again and so my initial distance zero but it goes down 500 meters. I'm not going to re-draw the cliff because it takes a lot of space upon my limited chalkboard.

So we know that the change in distance is equal to -500meters. And I’m still going to use the example where I do not just drop the ball or the penny or whatever I'm throwing off the cliff. I actually throw it straight up so it’s going to go up and slow down from gravity and then go to zero velocity and start accelerating downwards but you can even say decelerating in the other direction. But the initial velocity, vi=30m/s then of course we know that the acceleration is equal to -10m/s2 because acceleration gravity is always pulling downwards and towards the center of our planet.

If we want to figure out the final velocity, we could’ve just used the formula in the last video, vf2= vi2+2ad. But now, want I want to do is I’m going to use the formula we’ve learned in the very last video to figure out how long does it take to get to the bottom off― to hit the ground?

So, let’s use that formula. We derived to that the change in distance is equal to the initial velocity times time plus acceleration times squared over two. That’s initial velocity. So the change in distance is -500 is equal to the initial velocity. Well, that’s positive going upwards 30-meter per seconds, 30t― I’m not going to write the units right now because it just makes things― I'll ran out of space but you can re-do with the units and see that units do work out.

When square time, you just have to square time units etcetera although we’re solving for time, plus acceleration. Acceleration is -10. We’re going to divide it by two, right? So it’s -5t2 so if -500 is equal to 30t+ (-5), we could have said -5t2 and get rid of this plus.

So at first you’ll say “There's the t to the first, t to second. How do I solve this?” And hopefully, you’ve taken Algebra 2 or Algebra 1 in some place and you remember how to solve this otherwise you're about to learn the quadratic equation although I recommend you go back and you learn about factoring in the quadratic equation which there are videos on that I've put on YouTube so I hope you watch those first if you don’t remember.

But what we can do, let’s put these two right terms on the left-hand side and then we use the quadratic equation in solving. It sounded it’s easy to factor. So, we get the 5t2-30t-500=0. I just took these terms and put them on the left side. And we could divide both sides by five just to simplify things so we get t2-60-100=0. All right, I could do that because 0 divide it by five is just five so it just cleaned it up a little bit. So let’s use a quadratic equation for those of us who need a refresher I'll write it down.

So, the roots of any quadratic― in this case it’s t― we’re solving for― t will equal -b±√b2-4ac over 2a where a is the coefficient on this term, b is the coefficient on this term, -6 and c is the constant so it is -100. So, let’s just solve.

So, we get t=-b so, negative― this term. Well, this tem is -6. So if we make it negative, it becomes +6 so it become six plus or minus the square root of b squared, so it’s minus six squared, 36, minus four times a. The coefficient on a here and that’s just one times one for ac, c is a constant term minus 100, -4×1×-100×-100 all of that over 2a. A is one again so all of that is over 2.]

That is equals 6 plus or minus the square root—this is minus four times minus 100, so this becomes plus so this becomes 36+400 so 6 ±436/2 and this is not a clean number. And if you type it into a calculator, ― something on the order of about 20.9, we can just say approximately 21, and you might want to get the exact number if you’re actually doing this on a test or trying to sense send something to mars. But for all purposes, I think you get at the point.

So I’ll say it approximately now because we’re going to be a little off but just to have clean numbers. This is approximately 21, it is a little 20.9. We’ll say 6±― let me just write 20.9/2 so t approximately six. So, let me ask you a question, if I do 6-20.9, what do I get? I get a negative number. And does a negative time make sense? No, it does not. That means it’s somehow in the past― I don’t want to get a little philosophical but the negative time in this context will not make sense

So really we can just stick to the plus, Right? Because 6-20 is negative so 6 will just― there’s only one time that solves this in a meaningful way. So time is approximately equal to 6+20.9 so that’s 26.9/2 and that equals what? 13.45 seconds. Time is equal to―. That’s interesting.

I think if you remember way back four or five videos ago when we first did this problem, we just dropped the penny straight from the height. In that problem, I gave you the time. I set it to 10 seconds to hit the ground and we worked backwards to figure out that the cliff was 500-meters high.

Now, if you’re at the top of a 500-meter cliff or building and you dropped something that has no air resistance like a penny, it has a very little air resistance, it would take 10 seconds to reach the ground assuming all of our assumptions about gravity and whatever. But if you were to throw the penny straight up off the edge of the cliff at 30-meters per second right here, it’s going to take 13.5, roughly 13 and a half seconds to reach the ground.

So, it takes a little bit longer and that should make sense because― you should let me― I have time to draw a picture. In the first case, I just took the penny and its motion just went straight down. In the second case, I took the penny. It first went up and then it went down, right? So, it had all the time where it went up and then it kind of went down on a longer distance so it makes sense. This time it is 10-seconds while this time it was 13.45 seconds.

So, you can kind of say that it took― well, you actually can't say that. I don’t want to get too involved. Hopefully, this makes sense to you. If you got a smaller number here, you would have gone and checked your work because why would it take less time when I throw the object straight up.

So hopefully, that gave you a little bit of more intuition and you really do have a lot in your arsenal now, all of the equations. And hopefully, the intuition you need to solve basic projectile problems.

I'll now do probably a couple of videos where I just took a bunch of problems. It just really, really drives the point home and then I'll expand these problems to two dimensions in angles. And before we get there you might want to refresh your trigonometry. I'll see you soon.

Transcription by: Scribe4you Transcription Services