Ch 02: 1D Motion (Kinematics)See all chapters

Sections | |||
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Intro to Motion (Kinematics) | 67 mins | 0 completed | Learn |

Motion with Multiple Parts | 53 mins | 0 completed | Learn |

Meet/Catch Problems | 23 mins | 0 completed | Learn |

Vertical Motion | 48 mins | 0 completed | Learn |

Concept #1: Vertical Motion & Free Fall

**Transcript**

Hey guys, so all the problems that we've looked at so far had an object moving on the horizontal axis maybe this way really this way or even at an angle but it was always on the floor. Now we are going to look at vertical motion which is going to be motion the Y axis with an object either going up or down. So let's check it out so vertical motion the first thing is that instead of moving the X axis we'll be moving in the Y axis so instead of having things like delta X and V X we're going to have things like delta Y and V Y. Now freefall is the first thing we'll cover, freefall is what happens when the only force acting in an object is gravity so if the only force that acts on an object is gravity we say that the object is in freefall. Gravity is a force so the force of gravity can be represented with F G there's a few other representations but we'll stick with that for now, it's in freefall you can think of it as the object moves freely freefall freely through space without the influence of any other forces any forces other than gravity, so only gravity. So for example if we have an object here and it's going down and the earth is over here somewhere so this object is being pulled down by gravity then this object is in freefall because gravity is the only force right, now if as the object is falling or as the object is being pulled down I'm holding on to it on a string there's a little bit tension there then this is not free fall, not free fall. Now freefall the term freefall is kind a little bit misleading potential because even though it says freefall, freefall also applies to objects that are going up so it actually has nothing to do with going up or going down it just means that 're going in the horizontal either up or down and the only force is gravity so if gravity is the only force you are in freefall. For example let's say this object here the only force is gravity F G but it's actually moving up is this freefall? Yes because it meets the criteria for freefall which is that the only force acting on it is gravity. So because the only force is gravity things simplify a little bit but to simplify them even further we're going to ignore anything that could affect its acceleration in the Y axis we're going to ignore things like air resistance surface area right and because of that simplification all the objects will have the same freefall acceleration they're all going to fall at the same rate so it used to be thought thousands of years ago that if well not thousands hundreds of years ago that if you released two objects of different masses they would always hit the floor at different after different times and in fact they fall at the same rate as long as you ignore air resistance right so which is what we're going to do here. Freefall ignores air resistance so all the objects will fall at the same rate regardless of their mass and they're going to fall with an acceleration of G and again G is the freefall acceleration that happens as a result of gravity. G is an acceleration and F G is a force, so when I say gravity I'm talking about the force of gravity if I'm talking about little G I have to say the acceleration due to gravity or the freefall acceleration thou very often these words are used sot of interchangeably there is a difference between them anyway the acceleration in the Y axis in these problems will always be G and G on earth is 9.8 different planets or in the moon or whatever you are going to have a different G. Now the question is it 9.8 positive or negative it depends it could be either positive or negative so I'm going to put a plus or minus here, G is the magnitude of the acceleration due to gravity is a 9.8 depending on a situation it's going to be 9.8 positive or negative because if you remember positive or negative in physics only has to do with direction. So you could think of G as being always 9.8 that's the number but you could have to stick a negative in front of it depending on the direction

Let's do an example real quick and see how that works so it says you release an object from rest from the top of 100 metre tall building so here's the building you're going to release it from here and it's going to fall and goes to here so this is initial and this is final. This is 100 meters right here alright and I want to know the speed we'll have just before hitting the ground so it's released from rest so the initial velocity is 0 and the final velocity is what we want to know what is V final over here this is still motion problem so I need just like any other motion problem I need to know three out of five variables to be able to solve this and I'm going to have three equations to work with here is my three U A M equations the three equations of motion the only difference here is that the V's are now not V X's they are V Y's and instead of delta X I'm going to have delta Y and the acceleration will be an acceleration in the Y axis which will always be G. So we no longer actually have five variables we only have four because G is going to be it's going to be constant sorry. I have one variable here two variables here I need three more lets list them and it's the delta Y between initial and final it's the time that it takes between the two and whatever acceleration you're going to have, now notice that I wrote delta Y not delta X the acceleration here is going to be gravity I'm not going to put a sign just yet and if I'm falling from a building of 100 meters my vertical displacement is 100 right I'm going from the beginning to the end so the gap that I cover is a 100. Now this 100 is going down because I'm falling down gravity is always down right this velocity right before I hit the ground is down because you're falling as well and this has no sign and time is no direction right. What we're going to do is we're going to say that going up is positive you have to determine whether going up is positive or going down is positive and for this one I'm going to say that going up is positive. If you look at this problem delta Y is going down if up is positive then down has to be negative, G is going down so if up is positive G has to be negative so and then this velocity here will be a negative as well. So now all we have to do is solve notice that I have one, two this is going to be -9.8 again G is 9.8 but my acceleration is negative G, so its -9.8 it's not that G is negative G gets a negative tacked in front of it so I know three things here's all the stuff I know one two three and I'm looking for V so I can solve for V using one of our equations the second equation works and it's V squared equals V initial squared plus 2 A delta X. This is the the way that you were used to using the equation right here but we're going to make some changes I'm going to say this is actually V Y, V initial Y 2 acceleration is G or negative G and delta Y. So all I've done is kind of rewrite that equation you could have already written it this way the second way here if you want doesn't matter as long as you eventually plug in the right numbers. The initial velocity in the Y axis is 0 so my final velocity will be the square root and the Y axis will be the square root of 2. -9.8, -100 so it's the square root of this which gives me I have it here 44.3 meters per second I left a space here in front of the number so I can put a sign is this going to be positive or negative. Now once you do a square root you're always going to get a positive number out of a square root you may not remember this but technically the square root of 9 is not 3 it's plus or minus 3 because you could have to manually insert a 3 in there and that's what we're going to do you look at the diagram and then you realize that this is going down going up is positive so going down has to be negative so I'm going to force a negative in front of this guy and that's because it is going down so that's my final answer that's the final velocity the same thing that we always did write your variables your five variables you need to identify three out of five pick an equation solve the only difference is that the acceleration is always gravity now we're moving in the Y axis right so let's talk more about the direction of G and the sign of G. There's often some confusion about that the direction of G is always down right technically it's towards the earth but in most of these problems the earth is going to be below you so it's just going to be down but the sign so the direction is always down but the sign meaning positive or negative will depend on your frame of reference frame of reference remember means how you determine the direction of positive what you determine the direction of positive to be if you look here I chose that going up is positive but I could have chosen this a little bit differently.

So let's check out basically all the possibilities you have so let's say I decided going up is the direction of positive and here I'm going to say that going down is the direction of positive so going up is the direction of positive let's look at this first and what happens if you are going down an object is going down. Well if you're going down that means that the direction of your motion is down and that means the direction of your velocity is down the direction of your delta Y is down as well these two by the way have always had the same direction right if I'm moving to the right that's the direction of my delta X and the direction of my V if I'm going down that's the direction of my delta Y and the direction of V. Acceleration is gravity and gravity is always down now let's look at signs going up is positive so going down is negative all these guys are going down so they're all negative, so this is if you're going down. Now here you're going up so let's put a diagram there so you're go up and that means that the direction of your velocity is up because you're going up and these two are always the same so you're going up so you get up here but acceleration is gravity so this is going down I said going up is positive this guy's up up so its positive positive this guy's down so it's negative, so that's how that works. Here if I decide for whatever reason that I want to say that the direction of positive is going down in this case if I'm going down it means my velocity is down my delta Y is down and gravity is always down but now I said going down is positive so I have positive positive positive if you're going down sometimes it's a good idea to say that going down is positive why because instead of having three negatives I have three positives and that's better yet you don't have to deal with negatives which can be kind of annoying. If I'm going up going up means my V is up so my delta Y is up but A is always down so going down is positive so going up is negative negative going down is positive, so are you supposed to memorise all these? No you're supposed to just understand it depending how how you choose your direction of going up I'm sorry the direction of positive is going up positive is going down positive then you have to pick the signs accordingly for your V, your delta Y and your A. So there's several ways you can do this some professors will do it that if you're going up we're going to say that going up is positive and some of them will say that if you''re going down we're going to say that going down is positive and there are some advantages to that but I think that the way that generates the least confusion is if we decide right here if we decide to take the direction of positive to always be up if you do the same thing over and over it's a little bit easier then your A Y is always going to be down right and if going up is positive then your A Y will always be negative so A Y will always be -G in other words - 9.8 again it's not that G is -9.8 G is 9.8 but A Y is -G. So long story short if you stick to this your A Y will always be -9.8 and that's what I'm going to do for all these problems so that I have this consistent way of solving this alright so now I want you guys to try this practice problem and let's see if you can get it.

Practice: An object is launched up and reaches its maximum height in 10 s. Find its launch speed.

Concept #2: Multiple Parts & Symmetry

**Transcript**

Hey guys, so now we're going to talk about vertical motion problems that have multiple parts. Let's get started. So similar to what we did in vertical in horizontal motion when we had multiple parts, we're basically going to break up the problem into tiny parts and then piece it all together. So when we have multiple intervals in vertical motion we're going to first draw a complete interval diagram and you might remember me drawing a lot of these it's when you go from A to B and then B to C. Basically we're going to draw the path of the motion I'm going to show you that in just a second then you have to select the appropriate interval. So now that you draw a complete interval diagram you have to decide to find what I'm looking for am I going to work from A to B or is it B to C? Once you do this you have to determine the direction of positive for that interval. Now in this case I'm always going to say that going up is positive but you might solve it a little different. You might actually sometimes going down as positive or maybe your professor does that way you're more used to doing it that way. Once you've picked the interval that you're working with you're going to select the appropriate equations. It's very similar to what we've used to do and then obviously you're going to solve that equation. So let's do an example here. You throw an object with 40 meters per second from the top of a 100 meter building so let's check this out. Here's the 100 meter building and you're going to toss an object up. I'm going to call this point A so I'm going to say that VA is 40. So this object's going to go up, reach its maximum height and then start falling and eventually hit the ground so I'm going to call this point B, I'm going to call this point right here C the significance of point C is that it's back at the same height from A we'll talk about that and I'll call this point D over here. So when I throw something up obviously it goes with velocity 40, that 40 is going to be positive because I'm going to say that going up is positive. That's this step right here but I'm just going to do that for the whole problem. When I get to B, the velocity a point B is just zero because it's the maximum height. When I start coming back down, this guy will have some velocity here and this velocity will be negative and then V, the velocity a point D, so let me just right here that VC will be negative and the velocity at point D will be negative as well. Remember the velocity at the bottom is never zero so if they ask you what is the velocity right before it hit the ground, it's not zero because it's right before it hit the ground. So it does have a velocity. So that's kind of what it looks like just one comment even though I drew this going up and then coming back down off to the side, obviously doesn't mean that this object actually like curves this way it's just that if I drew it up and down along the same line it'd be a mess to see. So let me clean this up it's just going straight up and straight down.

So now let's see what we're looking for. We're looking for how long will it take in seconds, so we're looking for delta T, to reach the maximum height? So this is asking us for delta T, you have to figure out what variable you are using right. So they're looking for the time to reach the maximum height, maximum height is point B. So it's the time to go from A to B. So the very first question is asking what is the time from A to B? Notice the instructions, you're supposed to draw everything this is already done and I already figured out the direction of positive it's going to be going up, now I have to for every one of these questions pick an interval and an equation. If I'm looking for the time from A to B, it's obvious that the only interval I can use is the interval from A to B. So what I'm going to do is I'm going to draw A to B over here and in doing that I'm kind of extracting that piece of the problem and moving it over here to deal with just A to B. Forget about everything else, all I have to worry about right now is A to B. The initial velocity here is the velocity at A which is 40 positive going up. The final velocity over here is the velocity of B which is zero. So that's two of my variables I'm supposed have five. The other three are between A and B and the time that it takes to go from A to B, that's what we're looking for. There is a delta Y here, how far up did you go, and the acceleration in the Y axis which is always going to be gravity. Now I'm saying that going up is positive so my AY is -G or -9.8 because it's going down. My delta Y is going up so it will be a positive number and again we already talked about how this 40 here is positive. So if you look at what I have I need three out of five I have one, two, three, I know three things, I'm looking for delta Y and I don't have, sorry I'm looking for delta T, and I don't have delta Y but that's okay that's just my ignored variable which will tell us which will hint at what equation to use. I'm going to use the second equation because that's the equation that doesn't have actually the third equation because it's the equation where the first equation because it's the equation that doesn't have delta Y in it. So V final equals V initial plus AT I can just plug in the numbers the final velocity is zero, the initial velocity is positive 40, gravity is the acceleration -9.8, T. So if I move things around I get that T is -40 divided by -9.8. Those two cancel and I get simply 4.08 seconds. 4.08 seconds is the time that it takes to go up. So that's that. Part B says how high above the ground will it reach, how high above the ground will it reach? Another way of asking that is what is its maximum height? Well in this problem maximum height is a combination of two pieces. It's the height of the building plus my height from A to B. So let me write that down. Maximum height will be the height of the building which is 100 meters plus the delta Y from A to B. So to be able to find the maximum height I have to find delta Y from A to B and then we'll be done. So let's do that, let's find delta Y from A to B. Once again the interval that I'm supposed to use is obvious it's clear it's delta Y from A to B which is actually this guy right here. That's what I'm looking for, let me make a little green arrow here. So that's what I'm looking for and I already have the interval drawn up here I actually already know four out of the five variables and remember if you have four out of five instead of three out of five you have two equations to pick from instead of just one equation to work with and I'm going to go and use the third equation because it's a little bit more straightforward. It says delta Y equals the initial velocity times time plus half of AT squared, that's the equation. Delta Y equals the initial velocity is 40, the time is the time to go from A to B so that is 4.08 plus half the acceleration which is -G so it's -9.8 and the time again 4.08 squared. I punch all of this in the calculator carefully and I did that and I got 81.6 meters and then the last thing is the fact that this is not the final answer you gotta plug in here so 81.6 goes here and your maximum height is actually 181.6 meters, that's your maximum height, that's the answer to part B.

So let's keep going, part C says how long will it take in seconds, so delta T, to come down to its original height? To come down meaning from B to its original height meaning to C. So what they're asking here is how long delta T to go from B to C? So once again the interval is clear from B to C, we don't have to figure that out. Let me write it here from B to C. So let me draw from B to C and what do we know? The initial velocity here is the velocity at B which is zero, the final velocity is the velocity at C which we don't have, let's leave it alone for now. I have delta Y from B to C. So if went up a height of 81.6, this is from here to here, this was 81.6 meters in 4.08 seconds so if I went up 81.6 then from B back to C it's going to be 81.6 because these points are at the same height. So my delta Y is going to be 81.6. The only difference is that going up is positive and I'm actually falling so it's going to be 81.6 negative. The acceleration is again gravity 9.8 but I'm going down and going down in this case is negative I'm always going to do it that way and delta T is exactly what we're looking for. I need three out of five to find T and I have three out of five, one, two, three. So this final velocity here didn't matter, the equation I can use to solve for this is the third one so the third equation which says delta Y equals V initial T plus half of AT squared. The initial velocity is zero, delta Y is -81.6 equals half -9.8 time squared. So if you move things around and you try to solve for time you get 2, -81.6 divided by -9.8 equals T squared therefore T is the square root of that mass in there and I did this and I got 4.08 seconds. Now where have we seen that number before? 4.08, we saw it right here. These numbers are the same and that is not a coincidence. That is not a coincidence. So the time that it took to go up is the same time that it took to come down to the same height. So we drop 81.6 meters and it took the same 4.08 seconds. This is the only time I'm actually going to solve for this because from now on you just going to know that whenever two points are at the same height the time that it takes to go up is the same as the time that it takes to go down. That is called symmetry and I'll get back to that in just a second let's now first solve for part D.

Part D says what velocity will it have at its original height when it's coming back down? Coming back down at the original height is point C so what this question is asking is, let's fit it in here, what is the velocity at point C? And the velocity at point C is this guy right here. This is the velocity at point C that I want to know. So a good interval to use would be the interval from B to C and that's exactly what we're going to do. We can use the first equation knock this out really fast. So the first equation says V final equals V initial plus AT. The initial velocity from B to C is zero because it's my highest point and all I have to do is plug the stuff in here. Gravity will be -9.8 our acceleration is -9.8 and the time is 4.08 seconds. If you multiply these two you get -40. -40 meters per second and again where did I see this number before? Where did I see this -40 before? Well it's the same as my initial velocity over here it's the same as my initial velocity so there's another thing that repeats which is the fact that if I go up with 40 I come back down with 40, the only difference is that this 40 is negative because it's now coming back down. Again that's because these points are at the same height, they're symmetric. So we're not going to do this again you're just going to know that you don't have to worry about that so let me kind of talk about that over here. If two points have the same height, for example in this case we have that the Y position of A is the same as the Y position of C they are symmetric or motion between those two points is symmetric. What matters is not the word but the conclusion what that means is that the time to come down would be the same as a time that it took to go up and that the velocity coming down would be the same as the velocity going up except that it has a different sign because it has different direction. So that's that. That's going to be very useful in these kinds of problems and when we get into projectile motion as well. So just remember that point about symmetry and you won't have to solve as many things. Let's do part E over here and wrap this up. Find the total time of travel. So the total time of travel I'm going to write this as TT or delta TT is basically the time for the three segments here. You can think of it as the time between, whoops, you can think of it as the time between A and B plus the time between B and C plus the time between C and D and AB is 4.08, BC is 4.08 so the only one I'm missing is this one. So one way to go about this is all I have to do is go find the time for the last the third segment this part over here and then I'm good to go but what I want to tell you is that it's actually easier to do it a little bit different. Instead of finding this time here it's always better to work with intervals when you're picking your intervals it's always better to use the interval at the highest point. That's because the initial velocity or the final velocity one of the two will be zero. So whenever possible you want to include B what I here called B or the highest point in the interval of choice to make equations easier. So it's actually easier to find a time from B to D. So what I'm going to say as a general rule whenever you're looking for the total time you're just going to think of it as the time to go up which in this case is A to B plus the entire time to go down which in this case it's B to D and notice that both of these include the highest point and that makes your life easier. No matter how many points there are along the way I'm just going to say one all the way up and one all the way down and that's going to be easier and here we already know this so all we have to do is find the time to go down so that's what we're going to do. Find the time to go down which is the time from B to D. So if I'm looking for the time to go down I have to use obviously interval from B to D if I'm going to do it this way. So notice that I picked an interval that wasn't the most obvious and sometimes this is going to be helpful. So the initial velocity here is the velocity at B which is zero, the final velocity is the velocity at D which we don't seem to care for yet, what else do we know here? There's three more variables delta Y, the time that's actually what we're looking for and the acceleration. The acceleration just negative gravity, why? Because going up is positive gravity is going down so gravity is negative. My delta Y is negative as well because it's also going down, I'm falling and this final velocity here will be negative as well because I'm falling. Now if I'm looking for T I have to know three things. So far I know A, the initial velocity there has to be something else and what I know here if you look closely is you know your delta Y. Now your delta Y is not 100, it's not 81 so it's not 100 it's not 81, your delta Y is the whole thing. You're falling the whole thing so your delta Y is 181.6. The only thing is that that's a negative number because you're going down. Now I know three out of five, I can solve for this. So let's do that real quick. Notice that V final is your ignored variable which hints at the fact that we're going to be using the third equation, determines the choice of equations. So delta Y equals VOT plus half of AT squared, is my third equation. The initial velocity is zero and I'm looking for time so I just have to move some things out of the way. -181.6 equals half -9.8 T squared. So if you move all the stuff around the negatives cancel just do that very careful, you get that T is the square root of some number and if you do all that I have it here T will be 6.09 seconds. Now if you try to make sense out of this answer you can because it took 4.08 to go up, 4.08 to come down this piece, so to fall the entire thing you should take more than 4.08 and it does take 6 something. So that answer kind of makes sense along with the other ones and then the last thing I have to do is just combine all the T's. So the total T, T total or delta T total is going to be T up plus T down. T up is 4.08 and the complete T down is what we just got, 6.09. If you combine the two you get a 10.17 seconds. 10.17 seconds, that is our complete time to go from the very beginning, up and back down. 10.17 seconds. The last question is asking for, part F, is asking for the final velocity or it's velocity just before hitting the ground. So it's asking for VD.

Now there's two ways I can do this. I can find, I know the velocity at point C is 40 so maybe I can just go from C to D and write an equation and figure out of the velocity at point D but once again we want to include the highest point as much as possible. So what I'm going to do instead is go from B to D. That's because one of my velocities will be negative and the equations are simpler and B to D is actually what we did right here. So this final velocity that I want is actually this one right here. And once again I'll be able to just use the very first equation because I have a lot of information about that interval already so it's going to be V final equals V initial plus AT. So V final is VD, V initial is VB plus acceleration which would be gravity negative so -G and then times the time. The initial velocity is zero and the time that it takes is the time from B to D. Now every time you plug in a number on an interval on an equation for an interval, it has to be the number that belongs to that interval so we're talking about B to D so this time is the time between B to D and the reason why I'm stressing that is because there's a bunch of times out there, you have 4, you have a 6 and you have a 10. Here it's the time from B to D and the time from B to D was a 6. The time from B to D is 6, 6.09 and then you multiply these two numbers I have it here and you get a final velocity of actually don't have it here but you're going to find of a velocity of, I'm going to approximate this, of about -60. So I got a negative, does that make sense? It does. I am expecting a negative number because this velocity is going down so I know this was a really long question but you gotta know this and this basically covers all the possibilities of what you might be asked everytime you throw something up and it comes back to a height below the original height so it dips lower. If you understand this it's also massively helpful when you get to projectile motion. So that's it for this one.

Concept #3: Solving Multiple Parts in One Step

**Transcript**

Hey guys, so in this video I want to show you two special variations of vertical motion problems let's get started. So the first thing I want to talk about is the single step solution, now in solving vertical motion problems I've shown you how we can break them up into multiple parts and then bring the parts back together so for example we go from A to B and then from B to C. But sometimes it's going to be possible or perhaps even easier or faster or better however you want to look at it but you are going to be able solve part of the problem using a single step. So instead of going A B, B C you can go A straight to C or A straight to D. Let's do an example here. So you throw an object directly up with 30 meters per second from the top of a 40 meter building, so some like this and it start's from here you toss this thing up so this is point A and you throw it with a V A of 30 going up obviously this thing goes up to its maximum height continues all the way until it hits the floor. I'm going to call this point B this point C what's special about C is that it's at the same height as A so they are symmetric though in this question won't matter. At this point here my velocity at point B is 0 and remember the velocity at point C is because of symmetry is 30 but 30 negative because it's going down and then here point D there is a velocity at point D as well and that's what we're looking for in this problem what is the velocity at point D. But we want to do this using a single step so calculate its velocity use positive or negative to indicate direction now this is going down I'm going say that going up is positive which means that this guy should be down, so when I get an answer it should be negative and then we'll talk about that once we get there. So I want to do this using a single step in other words I want to know V D but going from A all the way to D why? Because D is my target I want some information at point D and A is my given it's my known because I know a lot of information about A I know how high A is and I know the initial velocity at A. We are going to do this in one single step which is possible so it's a motion problem which means I have five variables V initial V final acceleration delta Y and delta T and I need three out of five like always. If I'm going from A to D, V initial is V A which is 30 positive because it's going up V final is V D which is what we're looking for. Acceleration is gravity but gravity is going down so this is going to be negative 9.8 because it's meters per second squared. Now I'm looking for this I don't know how long it takes to do all this what about delta Y if you look around you're looking for V D you have one, two variables you need a third one it's going to be your delta Y. Now this is a little tricky at first or could be a little tricky at first I'm going from A to D. So forget what happened between A and D, what matters is that I started at that A and then I'm going to D. So effectively my motion looks like this yes I went up but then I went back down effectively my motion is I started at A and then I ended at D so my delta Y from A to D is this height right here this is 40 but it's going down so it's -40 and this is the only part of this in my opinion that's potentially confusing everything else pretty straightforward. Now I know three things and I can solve for the final velocity your time is your ignored variable we don't need time we don't want time.

Which tells that us we're supposed to use the second equation it doesn't have time so V final squared equals V initial squared plus 2 A delta X or in this case delta Y. The velocity final is what we were looking for so I just have to plug in all the other numbers, so this is 30 squared plus 2 -9.8 -40 and if you do all this you get that v is the square root of I got it here 1684 and that gives you about 41 meters per second. Now when you get this out of a square root you get a positive but remember the square root of a number like squared of 9 is really technically plus or minus 3 right so this is plus or minus 41 you're supposed to realize that it's that it is negative so you add a negative there. You're supposed to force that negative there and this is your final answer in single step. So sometimes you can do that now if a problem is going to ask you to find the time that it takes to go to the top time to fall so all the numbers in between you might as well break it up and that makes it a little easier and work with one interval at each point but if you just have to find one number from beginning to end you can do it like this. I just want to show you that you could do that, the other thing I want to show in this video is I want to talk about objects that are dropped from moving vehicles so you're in a car and you release a ball or whatever or maybe even you're walking and then you drop something right so an object that is dropped or launched are called projectiles whether it's from a moving vehicle or if you're standing if you throw something if you drop something and it starts moving it's a projectile but if an object is dropped or launched from a moving vehicle again it could be you you're walking you drop your cell phone right. What's going to happen is that the object will its will borrow so it borrows the vehicles velocity.

So if you're walking this way with 5 meters per second and you drop your cell phone, your cell phone will borrow your 5 meters per second this way. So it won't really fall straight down it will fall while going that way with 5 meters per second right. So let's do one example here it says you're inside an air balloon that is moving down so let me try to draw a little crappy air balloon, so that's terrible. So you're here inside of the air balloon and this air balloon is moving down with the velocity of it's moving down with the velocity of 10 meters per second when it is at a height of 30 meters and it says here at that point you drop your cell phone so here's the floor and I want to know the phone's velocity just before it hits the ground so here's your phone and you drop it and I want to know so obviously it's going to go from here to here and I want to know what is the velocity just before it hits the ground. Now the phone's initial velocity is not going to be 0 you didn't give it any initial velocity but it borrowed the 10 meters per second of the balloon so this is basically its going to be 10 this way this is the same thing as if you were on top of a building that is 30 meters high and instead of dropping your cell phone you actually through it with an initial velocity of 10 which in this case is -10 because it's going down. So this situation is exactly the same, now we can solve this I go from initial to final and I want to know the final velocity right same case here what we know between this interval I know that the delta Y is negative 30 why? because going up is positive this thing is going down so this is going down so it's negative this velocity is going to be going down so it's going to be negative as well and what else do I know here I know A Y is gravity 9.8 and it's also negative because it's going down I don't know time but I don't need time so time is my ignored variable because I already know three things so we're good to go and once again we're going to use our second equation final velocity equals initial velocity plus 2 A delta X or 2 A delta Y and if we plug in all of our numbers here final velocity is what we're looking for. I get that this is the square root of that's a -10 though it doesn't matter because it's squared anyway and if you do all of this I have it here you get 26.2 meters per second again this is either positive or negative because it's coming out of a square root in this case you analyse the situation determine that it's a negative because it's going down and I said that going up was positive so this is your final velocity. So again when you have objects leaving a moving vehicle they borrow the speed of the vehicle and then you can almost always convert it into some sort of problem where someone's throwing something. So now I want you guys to practice one this question actually sort of combines both things we talked about in this video it talks about an object leaving a moving vehicle an air balloon but that is now going up and it talks about you trying to solve this I want you guys to solve this with one step so it's a combination of the two things we wrote two things we talked about doing one step and with the object moving from leaving a moving vehicle so give this a shot and hopefully you get it.

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A rock is dropped from a high cliff (with an initial velocity of zero). What is the velocity of the rock 1.8 seconds after its release?
A) 9.8 m/s
B) 17.6 m/s
C) 8.8 m/s
D) 35.3 m/s
E) 15.9 m/s

A rock is dropped from a high cliff (with an initial velocity of zero). What is the velocity of the 3.6 seconds after its release?
A) 9.8 m/s
B) 17.6 m/s
C) 63.5 m/s
D) 35.3 m/s
E) 127.0 m/s

A rock is dropped from a high cliff (with an initial velocity of zero). Through what distance does the rock fall in the first 1.8 seconds of its flight?
A) 31.8 m
B) 15.9 m
C) 17.6 m
D) 9.8 m
E) 8.8 m

A rock is dropped from a high cliff (with an initial velocity of zero). Through what distance does the rock fall in the first 3.6 seconds of its flight?
A) 17.6 m
B) 127.0 m
C) 31.8 m
D) 63.5 m
E) 9.8 m

A ball is thrown into the air, directly upward. If it feels no air resistance, which of the following statements is true:
a) The acceleration of the ball points downward while the ball travels upwards, and upward while the ball falls.
b) The acceleration of the ball points upward while the ball travels upwards, and upward while the ball balls.
c) The acceleration of the ball points downward while the ball travels upwards, and downward while the ball falls.
d) The acceleration of the ball points downward while the ball travels upwards, and upward while the ball falls.

An object is thrown vertically into the air at 5 m/s. What is the maximum height the object will reach?

An object is thrown vertically into the air at 20 m/s. What will the object’s speed be after 0.5 s?

An object is thrown vertically into the air at 15 m/s. How long will it take the object to be moving downwards with a speed of 8 m/s?

While visiting New York City, you want to see how long it would take a penny to fall from the top of the Empire State Building. If the observation deck of the building (on the 82nd floor) is 320 m high, how long would it take the penny to hit the ground?

An object is dropped from an arbitrary height. How much further does the object fall in the 2nd second than it does in its 1st second of free fall?

A ball is thrown downward with a large starting velocity. Will this accelerate more rapidly than one that is just dropped at the same time from the same height?
A) No, because it starts faster it will later accelerate less rapidly.
B) No, freefall acceleration is always the same, no matter what the initial velocity.
C) Yes, a ball that already has a higher velocity will also accelerate at a higher rate.
D) Yes, but only of the thrown ball is also heave than the dropped ball.

A ball is thrown straight upward and then returns to the earth. Does the acceleration change during this motion?
A) The acceleration first decreases and then increases.
B) During the motion the acceleration never changes.
C) The acceleration is larger going down than up.
D) The acceleration first increases and then decreases.

A ball is thrown upward from the ground with and initial speed of 41 m/s; at the same instant, a ball is dropped from a building 37 m high. The acceleration of gravity is 9.8 m/s2 . After how long will the balls be at the same height?

A ball is thrown downward with a large starting velocity. Will this ball reach the ground sooner than one that is just dropped at the same time from the same height?A) Yes, because the average speed of the thrown ball is larger it will hit the ground first.B) No, because the acceleration is the same for both balls they will reach the ground at the same time.C) No, the maximum speed for falling balls is 9.6 m/s and both will hit the ground at the same time.D) Yes, but only if the thrown ball is also heavier than the dropped ball.

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