Projectile on an Incline
It’s time now to take a look at a special case for projectile motion where you are throwing it on an incline plane, say on a mountain. Yeah? So when we are throwing it on a mountain what’s different? Yeah, your surface is not flat but it’s at some angle. We call that an incline. Right? So now I want to assure you one thing though. What I want to assure you is nothing fundamentally can change here. A 2-d motion problem will be solved as 2 1-dimensional motion problems. Yeah? The only thing here is you have to see which way to break it down into two 1-dimensional problems. Now let me tell you what I mean by that. There are two ways to look at a ball thrown on an incline plane. One of them is this. If I keep the incline plane. Let’s say the incline plane is at some angle θ and I am throwing a ball on it. It’s going like that at some angle, at some other angle I have to choose right because θ is already taken. So let’s say the angle I throw it at is α. Then the ball will do something like that. Right? And go hit this somewhere. Now this is one way of looking at it but the same thing of throwing a projectile at an incline can be looked at like this also. What if I imagine that the ball is thrown? Right? Just like any other ball on the Earth. Nothing much is different except that on the way it meets an incline plane. So an incline plane exists somewhere over here. It meets that. Because I can also argue, right, very, very validly that till the ball hits the incline plane how does the ball know it’s going on an incline, or it’s flying on an incline plane. There isn’t any meaning to that statement. Now why are these two fundamentally different ways of looking at projectile motion on an incline plane? I will tell you why. So what is different between these two approaches? In one you think of the incline plane first and then imagine a ball being thrown on it and the other you imagine the ball being thrown. It so happens that it is an incline plane. Now this seems like a psychological difference right? Yeah it’s more to it than that. The way you solve the problem will differ. Why? Because there are basically two ways to solve this problem. Right? The problem of a ball on an incline plane. And one is to where you tilt your head, and the other is where you say ‘yeah I won’t tilt my head’. Just because it’s an incline plane why should I tilt my head along the incline plane and think. You will say gravity doesn’t care Gravity always pulls the ball down. Yeah? So I will just keep my head straight and it so happens that it’s an incline plane at some point, whatever happens I don’t care. So what I am trying to say is, in one you can choose your frame of reference to be that way along the incline plane and in the other you will choose it to be the usual old up and down itself. Does that make sense? So in one you are frame of reference will be tilted like this. Yeah? Along parallel and perpendicular to the incline plane and the other you should say I will still keep it my old way, which is up and down, or sorry, right and left. Yeah? So overall you will have a horizontal and vertical itself. Now let’s take this one first r. The old one itself and see what happens. Let’s say you want to ask the question how long will this ball take to hit the incline plane. Let’s say that’s the question you want to know the answer for. Will it be even different? Clearly it will be different. Yeah? Because if you throw it here it will take much longer to reach the ground. So it reaches here much before. Does it depend on the angle θ of the incline? It definitely does. Yeah? Because you can see it here itself. The more and more you make the angle θ the more and more, the shorter and shorter the time will be. Yeah all this is your intuition. Right? Let’s first do it with this approach which is basically not changing your frame of reference. Now what does that mean? For you that just means that a ball’s being thrown. An incline plane comes in the way somewhere. So all that matters is what are the x and y-coordinates of that point where the ball will hit the incline plane. So let’s say this distance is some x. Right? Then this distance will be what? It will be some y. Yeah? So this point will become y, x. Right? But are these x and y independent? No they are constrained. They are constrained by the fact that it’s on that incline which means this y is constrained to be y x tanθ right? Because y by x will be equal to tanθ. Makes sense. So now you’re constrained by this idea that your y and x are related. Then your coordinate system will become (x, x tanθ). You know what’s beautiful here in this case? You can forget that this incline plane exists. Doesn’t matter to you anymore. All you have now is the projectile that is going between the points, or rather passing through the points (x, x tanθ). What you are trying to find out is at what time t will its y-coordinate be x tanθ. Because you know who determines time? Only the vertical. So the advantage of this is more psychological because if you don’t like inclined planes in one step, I am giving you a way to remove the incline planes from your picture. From now the incline plane doesn’t matter at all. The incline plane only mattered to impose this constraint. That the y and x will be (x tanθ, x). You can solve the problem here. Let me show you. So now what matters to you? Only your y. Yeah? Only your y component matters to you for finding out the question of time. So what are your equations? Nothing is going to be different right? It’s our old problem. So I don’t even have to write but I will just write it for now. So your, we haven’t named the velocity. Initial velocity we will call it u as we always do. Yeah? So this, one important difference before you remove away your incline plane is what? That other extra angle Right? α will be over there. That’s the only thing you have to keep track of. And here because it’s given variables it might look complex but if it would have been given 15 and 30 you get 45 as your final angle. So this overall angle will be θ + α. That’s the only change. After that the incline plane is gone. So what will your equations be? Your horizontal velocity will be: u cos (θ + α). Your vertical will be: u sin (θ + α). Advantage – no incline plane. So for your time what matters is only your vertical component. Yeah? So what will your equations be? u sinα, sorry (θ + α). Yeah? Into t minus half g t2. Old original equation, that’s just the equation for the vertical motion. So I will write y-motion will be equal to x tanθ. That’s one equation. Yes. Now your x-motion will be u cos (θ + α). That’s the initial motion, initial velocity along the x. Does this change here? It doesn’t right? There is no incline, nothing else. Just going horizontally. No acceleration. Into time will be equal to your x. Yeah? Simple and straight. So these are your two equations. You have two equations, you have two variables. You want to know your t. You also don’t know your x. Correct? So the rest everything is constant, you notice right? Because your u is constant, your θ is a constant, your α is a constant and there is nothing else over here. Two equations, two variables you can solve. You will get the answer. Yeah? For some questions this might even be a good approach but now you are going to take a look at the other approach. The head tilting approach and see, like we told you, sometimes changing your point of view can become, can lead to a lot of simple answers. So all said and done, even now this is a lot of calculation. Right? At least somewhat of a calculation. What you will do is you will substitute depending on what you want. If you want time in this case you will substitute for x over here. Yeah? So u cos(θ + α) into t will come over here. And then you can compute your answer. But now let’s use our idea of a frame. Changing our frame and see what we can do.
So now we are going to take the approach that we are going to tilt our head. And let me draw the diagram again. So you have an incline. That’s where you begin. And it’s at an angle, θ, like we said you are throwing something at an angle, α, it’s going to go and hit somewhere over here. Yes? We are assuming that it’s on the incline. Otherwise we will make the incline longer so that it will hit. Now what we saying, what would it be like if I shift my axis? Instead of being horizontal and vertical to be parallel and perpendicular to the plane. What would that look like? So now, what’s the disadvantage of this? Yeah think about it now. Why am I so hesitant to change it that way? Because all through my journey one guy was always pointing down, guy or girl, yeah it doesn’t matter. Acceleration due to gravity. So which meant I did not have to take components of that. Yeah? If they were perfectly aligned along my axis, so one g’s component is always down, the other component had no acceleration at all. Made it very clean and simple. I am sacrificing that over here. So if I am sacrificing that simplicity I need to be rewarded with some other kind of a simplicity or symmetry or otherwise I won’t do. So we are experimenting. Let’s see. Let’s see if we are rewarded. So what’s the implication of this? You will take your components along these two directions. So your vertical velocity will be u sin α only. Horizontal will be u cosα. Now g also needs to be taken into components. So your g would have been somewhere over here always pointing down. Now your g will have two components. One parallel to the plane and one perpendicular. So the difference here will be that both motions will be uniformly accelerated motions whereas in the original where your axis was horizontal and vertical you know that the vertical one was accelerated but the horizontal was uniform motion. Yeah? So we are going from that model to a model where both those are, both the 1-d motions are accelerated. But what is remaining constant? Dividing two 2-d motions into two 1-d motions. That’s what we are doing throughout. We are just thinking how to do it better. That’s it. So what will these two axis be? This will be g into. Yeah? Do you want some time to think about whether it will be sin or cos. It will be sin okay? g sinθ, not α, g sinθ. This will be g cosθ. If you want a minute to think about why there is, let me just show you. At some point if you want to draw, take components, then here for a line like this, let this be g. Doesn’t matter right where I draw? Well I am drawing it here to give you a clear picture of it. You try and take components along this side and along this side. Right? So this is 90 right? Yeah. That angle is 90. So this will be 90 – θ, which means this will be θ. That’s all that is important. After that you can forget this diagram because from here I can translate to, this is θ. That’s enough for me. The remaining will, the other will be 90 – θ is very obvious. So I will erase all these bits over here. Great. So now you have g cosθ and g sinθ along these two directions. So your equations are going to look slightly different. Right? You will write your equation for y, yeah, and in this case your y will be, we will replace the words vertical and horizontal with perpendicular and parallel. So let’s call it perpendicular motion and x-motion will be parallel motion. Don’t put pen on paper and write these two equations. That anybody will do. Yeah? You can write these equations. It’s done. Your problem is solved okay? That was solved long ago in 2-d motion itself. Because projectiles only a special case. Why would we do something like projectiles separately because there is some symmetries to be unveiled over here? So you can put your pen and paper, now write the equation for x. Yeah? We will do that and show you need not have done that. To answer the question what time will the ball take to hit. Yeah would you believe me if I say it’s a mental problem? Yeah you can just do it in your mind. So, let’s see what happens. Why perpendicular? It’s going to be what? Your initial velocity is u sinα into some time t which is what you want to find. Minus half, g or, yeah g cosθ, because it’s the perpendicular motion the acceleration along the perpendicular is g cosθ. g cosθ t2 equals what? What is the final displacement along the perpendicular direction? From here to here if this is your axis, initial y value is zero, your final y value is also zero. Yeah? That’s the advantage over here. You know. Because you are thinking of it this way. Yeah? It’s going up and coming back down but along that plane its initial, it was on that plane, so initial coordinate, y-coordinate is zero. Finally also it is zero. So this will be your equation. What about the x? Let it have travelled some range R, then the value of x is that range. So you will be u cosα t. Very, very symmetric equation because now both are accelerated motions. g sinθ t2 because along this you will have a g sinθ that also pointing in this direction so it’s a negative g sinθ. Yeah? Hope I am not being too fast. You can pause and just think about the signs and all. I am assuming those will be easy for you. Equals range R which is positive. Yeah, this R is positive because you have taken this to be positive, this to be positive. This to be negative, this to be negative. So both the components of g are negative, as you can see. Right? So these are your two equations. How many variables? Yeah? This equation has only one variable actually. Because u is known, α is known, g is known, θ is known. t we will get directly from that equation. But without putting pen on paper. Because it provides us with some beauty. Let’s do it the other way. Let’s ask the question – what’s different between a projectile thrown on the land and a projectile thrown on an incline plane. As far as the vertical motion is concerned or in other words a perpendicular motion. The time of flight is determined only by the perpendicular component. Do you agree? Let me draw this out separately. If I have one thing like this and if I have one thing like this. Both these cases, the delta t is determined only by this component and this component respectively. Correct? You agree? Over here what was the acceleration acting in this component? g. Yeah over here what’s the acceleration acting on this component? g cosθ. Yeah? So if I had thrown a ball at α over here like this and now I am throwing a ball at α on an incline plane of θ. These are the only two differences. The rest all is going to be same. This is going to be u sinα, the velocity in this direction. Yeah? This is going to be u sinα here also. Doesn’t matter along this direction. What does not matter is the perpendicular component at all. It doesn’t matter at all, right. I am sorry. The parallel component. What’s happening along the horizontal line and the parallel direction here hardly matter. Because they don’t determine the time. The time is determined by the perpendicular. I am telling you all this to finally convince you psychologically that nothing is different between this diagram and this diagram except that you are replacing g with g cosθ. What does that mean for you? In this case you know what the answer for time of flight is. u sinθ to go up and u sinθ to come down. So overall 2u sinθ by g. You are throwing with α now so it’s just 2 sinα by g. Correct? 2u sinα by g is the time over here. That also I am not asking you to remember. Right? We did it before itself we told you, you can derive it in one second. Yeah? Because it takes u sinα by g to go up, u sinα by g to come down. You are done. What should be the only difference over here now that we are observing this symmetry? See this where problem solving becomes beautiful okay, it’s not a mechanical thing you keep doing but you are trying to observe symmetries and beauties. If you replace g over here with g cosθ, the angle of the incline plane, you must get the answer over here. Because you have shown that nothing else is different. So let’s say without putting any pen on paper you are saying it will be 2u sinα by g cosθ. Amount of calculation required zero. Right? Amount of thinking involved, a little bit. But now you know this next time you have to think you will be very, very fast. The insight you are using is, on an incline plane all that happens for the perpendicular component is g gets replaced with g cosθ. That’s it. So you have to go up here. Now let’s see if what we said is arriving at, you know are we getting the same answer mathematically as well. So let’s to be cleaner, let’s copy down this equation that we want and take it down there and solve it separately. Because that’s the only equation that we want right? So this is the equation that we want. Let’s take this with us and let’s see if our prediction made based on intuition is being matched over there. Okay? So the u sinθ, α into t part didn’t come with us. It chose not to so let’s write it down over here. u sinα into t. Yeah? We are sure this is the right equation right? Let’s go back and check. u sinα into t minus half g cosθ into t2. Yeah? I am being meticulous here because you can petrate again but in general these kind of things I only do in exams go wrong so I won’t do it now. So now let’s solve this it’s only going to take like a minute. What comes out now, t u sinα minus half into g cosθ equals zero. Into t equals zero. So 1 and t remains inside. Yeah? What this does imply is that t is equal to zero is a solution. It clearly is. Yeah at t is equal to zero also the ball has a displacement of from the perpendicular of zero only. So that’s not surprising. One of our solutions is t equal zero. It will always come. Even in our normal horizontal/vertical problem that came. Right? When you do that way you will get t is equal to zero and t equal to the overall time of flight is your answer. So the same thing is going to happen here. What’s it going to be now? You get t equals u sinα and if 2 has to go to the other side. So 2u sinα by g cosθ. Great. So of our observation of symmetry worked. We have mathematically verified it. We kind of verified it here as well right? We applied sure here but it’s good to check it with one more step. We have shown here that changing your point of view and looking at it. Changing your frame actually made a long problem because over here if you go back to our original case where we kept it without imagining an incline plane as a normal problem this is still good. You will get the answer but there is somewhat of a math involved. Yeah? But you have shown that it can be beautiful and symmetric and simple if you if you are to change your point of view. Now what will the range be? Yeah? You have got a time of flight for the incline plane by doing nothing. Just replacing g with g cosθ in your original formula. Now your range will be nothing much right? What’s your range? If you have an angle like this, that’s your range. Yeah? What’s the acceleration in that motion? Two 1-d motions. One you have already solved. Second 1-d motion you are solving. So its g sinθ is the acceleration in this direction. So what will your equation be? Yeah, what do you need here? The time that you just found out. So your equation will be what we wrote before itself. Let’s just refer to it again. Yeah, this equation is what we are going to use now. Range equals this quantity. So let’s write that down again. So we will have u cosα, because that’s the initial velocity. Time was, we just found out 2u sinα by, instead of g, g cosθ. So we write that down, all this down over here. u cosα, your t being 2u sinα by g cosθ minus half into g sinθ into t2. Again t2 will be 4u2 sin2α by g2 cos2θ. This is the part where you are going to feel a little like ‘oh my god’. Yeah? Nothing much right. This is just an expression that looks long. We can’t force nature to be always simple and symmetric and beautiful and come out to a small answer, right? We cannot. And anyway this looks more complex in comparison to what? To a horizontal motion your answer was u2 sin2θ by g, instead of that here the range, this distance covered along the incline plane is going to be given by this. Don’t remember this and all. It’s pointless. You know how to find the time. Yeah? That’s very easy. After that just calculate this 1-d motion. You are finding out the displacement after some time t. That you already know how to do right? In accelerated motion find displacement in 1-d motion, after some t you know what to do. This is exactly what we are doing here. Right? So this expression particularly I don’t know maybe we can simplify in some way. You can take t out common and write something but yeah once you are given intuitive numbers in problems you will be able to solve it easily. Now you could have done the same thing again. The same range problem also you could have solved using the approach without turning your head. So you have got your x value. Yeah? How do you calculate the range with your x? That’s your range. That length is your range. What you will do is you will write range equals x by cosθ, right? I am not using some big formula here okay, you know that range cosθ equal to range cosθ will be equal to x. Yeah? R cosθ equals x. We are just writing it in the other way. Range will be x by cosθ. So the moment you have calculated x here, and one advantage is if you want range, time, couple of things then doing this might actually help you. Depending on the problem think about it. Then you might find x and then use it divide by cosθ to get your range. Now I know one thing though. For the question of when is it farthest away from the incline plane? In other words finding this distance, right? For this problem surely tilting your head seems to be better. Yeah because I have this, because I forget the incline plane I will have no way to measure when does it become maximum? There will be some complexities here too. Not saying it’s impossible. Surely whatever you do in frame can be done in the other frame. Where is it simpler is the question. So this particular problem is much easier to solve when you tilt your head only. What will you do? Find out the vertical displacement. In this case the perpendicular displacement. I know I tend to use the word vertical/horizontal out of habit right? In incline plane what we mean is the perpendicular and the parallel. So what are you being to find here? The maximum perpendicular displacement. That’s your, this height right? Yeah, it’s not very accurate to call it height. So we will call it some H incline. So that we still remember our things correctly. So if takes 2u sinα by g cosθ to go that far, it will take u sinα by g cosθ to come here right? That half rule will still hold here. All the symmetries are true. Right? Because it takes 2u sinα by g cosθ, it will be half of that. Then all you have to do is write perpendicular equation for that right? Whenever I am giving you some time, it’s time for you to guess what I am going to say. So you have, and also time for myself to scroll the pages. So you have u sinα into your time which is, u sinα by g cosθ. Not 2u sinα by. Yeah? Minus half into g. That’s right. cosθ into t2, which is, u sinα by g cosθ, the whole square. What does this give you? Maximum displacement from incline. Maximum perpendicular displacement from incline. Again this doesn’t have to look beautiful because now you know the method. There could be some simplifying possible here but it’s not really necessary. This gives you that distance. So what we have done here is show you to answer the questions how long, how far and high from the incline. Yeah? Just like you did for the original flat case. Same angle – same plane case. And you know that this is not fundamentally very different. You have also observed that tilting your head gives you some very, very interesting ways to simplify the problem.
Now as a last step to this. Right? I want to address one kind of question that is asked very commonly. Right? And will show you that it’s even simpler than this. So if I throw something on a staircase, or say, from a staircase. Like this. And if I am asking questions like which step will it land on? Or which step will it hit and things like that the first thing and the last thing I want you to observe is that every staircase is defined by an incline plane. So it’s an inline plane problem. All you will find out is – so in these cases sometimes not tilting your head might help out. So you had two approaches for incline plane problems. Because if you have to find out which step it’s going to land on. Say it lands like this you throw and it’s going to land somewhere over there. Here let me use a different color. Let’s say your ball lands somewhere there. The answer is going to be, first step, second step, third step, right? That’s it. For that what should you find? The x-coordinate of where it hits the incline plane. So you can forget this staircase, calculate the, keep it as an incline plane. So you are looking at steps and steps of simplifying the problem, right? Because once you keep it as an incline plane, you can forget the incline itself. It’s just (x, y). Yeah? So you have x and x tanθ that the staircase will define for you. You will just take this as steps that are equal. In other words, this line equals this line, what will the angle be? Yeah? 45 degrees. Right? Has to be. Based like this for any other relationship given. If they say that this and that is twice, this is twice of that then this angle will be, I don’t care what angle it is, okay. But the tan of that angle will be 2. So that tanθ will be inverse 2 but you can do this and then because here also you will need x tanθ only. Your tanθ will be defined. So I am just showing you that whenever you see a staircase problem, draw an incline solid as an incline plane. Then answer whatever question you will be asked. Yeah, if you know the x-coordinate and if you know the length of each staircase, the horizontal length then it, your answer will just be x by that length, right? Which stair will it hit? Maybe if I take a number here and solve it you might get an example but that’s done. That’s done in our question bank. Go check it out. There will be problems where you will get perfect practice for this. It’s a very simple idea. So now I know you have very, very good grasp of incline plane problems, they are very, very easy.