Maximum Range in General
So when we saw about the maximum range of a projectile our reflex action answer was, throw it at 45 degrees. But we saw that it was only true when the body was thrown and caught at the same horizontal plane. When that was true, we knew that the answer was 45 degrees. Do we know anything else about any other style of throwing? Throw from here, catch here. Let’s say that you are standing on your terrace. Right? And somebody below wants you to throw the ball. And you want to throw it as far away as possible. You want the ball to land as far away on the ground as possible. Then, at what angle must you throw? Now you might still be tempted to say 45, like we discussed in the previous, couple of videos I think. So now if it have been 45, right, one of the arguments also you can give is this. Till the horizontal, 45 has the maximum range, therefore even after that, finally, it might have the maximum range. But, let me say if I think about that argument maybe not also, right. Maybe what you are saying is correct, maybe not. What I would argue is what if somebody that’s thrown at a lesser angle, say, is lesser over here. It’s you know, when it reaches the horizontal plane, it has had a smaller range. But after that what if it can overtake and land somewhere like that. I mean, doesn’t seem impossible, so the, what should I do? Forget all my assumptions and let math tell me what the answer is. That’s exactly what we are going to do. We are going to go on a slightly long journey, I know because I just did it sometime ago, to find out at what angle will you get the maximum range for a general case. We are not throwing and catching at the same height, but it can be anything at all. Let’s go and do that.
So let us say at the end of the journey, the vertical displacement is s, instead of zero. Why instead of zero? In the horizontal plane case what you did was you threw and caught, at the end of your journey, your vertical displacement, sy, was equal to zero. Right? And that was why it was a special case. Now if we assume it’s not going to be zero, but some s, right? Then what will that yield. Let’s watch. Let’s for now believe that, you know, it’s something like a cliff. So s can be either positive or negative. It’s being thrown like this. Yeah, do we care? We don’t. So you will draw it out. You are going to divide your velocity into its components. So let’s give some variables. Let’s do it from a general case. We can take some specific problem and solve it, but I prefer to do it with the variables because you can see a very meaningful answer. Let the acceleration be g downwards which means its minus g. Let the height that we are throwing from be, let’s not give the height as, let’s do it this way. Let’s say the final displacement is some s. Some positive s. What does it mean? If you are throwing from a height s, this value would have been, if this height had been some h, right, your s value would have been minus h. Because it’s downwards. If you are throwing at a wall, for example, and you want to reach as far away as possible on the wall, then, it would have been some plus value. So we will keep it as pluses, yeah? Do you realize what this means, right? And you’ll see it even more after we solve the problem. So instead of s being zero, it’s some value s itself. That’s the final displacement, general case. g is your acceleration. Let v be your velocity and let angle θ, be that θ such that you get your maximum range. And let R be your range and Rmax be your maximum range. So setup the problem. Let’s begin.
What’s our first equation? Nothing else is going to change. The first equation will look very similar. Vertical equation, Y equation let’s call it, will be equal to vsinθ into t minus half gt2 equals, that’s right, s. Right? We are using plus s here which means depending on the value of s it can either be thrown up or thrown down, doesn’t matter. It can even be thrown at zero. So it covers all cases. What about our x equation. x equation is very simple, right. What will it be? That’s right. It will be vcosθ, vcosθ into t equals your range. Yeah? This is what you want. We have written this equation in terms of a range because what we call range is this final x value that you cover, correct? And that is going to be equal to the horizontal velocity into the time. I am pretty sure you know this by now. Now it’s where you have to think a little bit because these equations will alright but based on what we want we have to play with these equations. Because honestly after this the physical intuition is over. Now from now on it’s the math intuition. Now let’s see what we must do. We want the range to be maximum. Right? Which means we want to do what? Range is a function of what here? As you keep changing this angle, right? If you change the angle to some other θ, some other θ, the final range is also going to keep varying. You know that for sure. At least from horizontal you 100% know that. So range is a function of θ. Then, if we can arrive at a function of rate and θ, range is a function of θ, then you can differentiate that function and equate that derivative to be zero. We’ll get the maximum range. This is the methodology. Now this same methodology that took one minute might take slightly longer when we do it but this is the basic methodology. You are trying to get a function of range and θ. Rest all are constants over here except for time, right? Now what this’ll allow you to do when you get a function of range and θ is, then you can differentiate and get the answer. So you see what we are trying to do here, that’s important. Then what must our first step be? Get rid of the variable that we don’t seem to need right now. And what is that – time. So how do we get rid of time? From here you can write time to be equal to R by vcosθ. Would you agree? Yeah, so now direct elimination of t. Plug this into the y equation and you are done. So let’s do that now. So your y-equation was vsinθ into t minus half gt2 equals s, right? So your t is going to become, like we just saw, R by vcosθ. So it’s going to become minus half into g into R2 by v2cos2θ equals s. You can follow along slowly or quickly and imagine doing the problem. You can even predict the future steps. You can try it out, see if you are getting the answer and then come back here also if you want to. It’s not like you have to watch what I am doing. It’s purely mathematical from here. There will be some ideas, but, I kind of encourage you to go from here, try and solve the problem and then come back and see if you got the right answer. So, I am going to assume that you are going to do that. And I am continuing now. So this is what you have. There’s some simplification you can do right. Let’s get rid of what can be got rid of. So you have this, you can write this as Rtanθ, do you agree? Minus g by 2, let’s keep all our constants together, g by 2v2 is a constant right? g is a constant, v is given in the question. Because you are trying to maximize. I know it’s obvious but you are trying to maximize the range for a given speed. You can turn the angle of the speed but that speed is fixed. So, minus g by 2v2 into R2 by cos2θ. Can I write that as R2sec2θ. Why? You’ll see now. Equals s. So this is an equation that is in the form that we want and what is that? It has only R and θ. No other variable. So we are tantalizingly close to what we want. Now, there is one problem though. If I want this in the form R equals some f of θ, then I must separate them right? What must I do? I must try and take the θ to the other side, but then, you know I have an R2 here, right? I didn’t copy the square over there, I tend to do it sometimes. So, separating R and θ is not going to be drivel, this is not going to be easy. It might take lot of effort. So can I work around that? Because sometimes we all get constrained because we think that to differentiate, my R should be on one side, f of θ should be on other side. Differentiate their function, equate it to zero. Yeah, when it’s difficult, there is another way to do it. And we have spoken about this in 1-d motion but you can think about it again over here and what is that? Just write this equation in terms of differentials. I will tell you what I mean by that. So you have this equation with you. Let’s take it down. Let’s write it again. If you have, your equation is Rtanθ minus. So you have this equation. Now, you can write this in differentials. What I mean by that is, you can just, what will happen is that this is a product of two functions. Yeah? So you will first differentiate this as respect to, you will take this as a product and differentiate with respect to every variable. So you will have dR for your R multiplied by tanθ. So what have we done here? We have applied the product rule where you are first differentiating R. So you get dR into tanθ plus, now you will not differentiate R, right? R is just that way. tanθ will become sec2θ. The derivative of tanθ is sec2θ but multiplied by another dθ. That will be the derivative of θ itself. So, or the differential form of θ itself. What we are, in one sense doing is differentiating with some third neutral variable, you can think of it that way. This is a mathematical technique. So what you doing, this part alone you focus pretty strongly. So this Rtanθ is considered a product of two functions, yeah? So first function R alone, you differentiate it on dR and multiply it by tanθ. Then you kept R constant and differentiated tanθ alone, and you got sec2θ.dθ. This exactly what you’re going to do throughout. So now if you continue, you’ve minus g by 2v2, yeah? First let’s now differentiate R2, so we have, first let’s differentiate R2 so 2R into dR, yeah? Multiply it by, the other thing is kept constant, so sec2θ minus, again, g by 2v2. Now R2 will be kept a constant, and you will differentiate sec2θ. And what is that going to be? Chain rule, so, 2 secθ will be the derivative of sec2θ. Yeah, it was 2 times secθ. Then you are to multiply with the derivative of secθ, which is, secθtanθ I think, yeah, secθtanθ. You can either look it up initiate or you can derive it by knowing the derivative of cosθ and write this as 1 by cosθ. I am just not doing it. I think I remember it correctly. I am pretty sure. So secθtanθ followed by a dθ for the θ itself. So this is a perfect example of chain rule being used, right? One loop of a chain rule, second one, third one. This whole thing must be equal to the derivative of s, but s is a constant right? For the given question it’s going to be a constant. At what height you’re throwing is not going to change as the ball is being moved. Correct? So, it’s a constant, so the derivative of that is going to be equal to zero. This made it so that we can avoid the step of separating the two variables. Now what’s beautiful here is, we can let’s say, divide the whole equation by dθ. Now I know I am using, vague words right, how can you divide by dθ? In its differential form, they kind of behave as if they’re a quotient, you will understand much more of this when you learn calculus next year. But trust me, you can divide the whole thing by dθ. So you will get, dR by dθ over here, so your dθ here will cancel and you will get dR by dθ over here as well. Right? But this is where the moment of understanding comes from our physics. If the range is maximum, then this value, dR by dθ must be equal to zero. No matter in which equation, in which form you write it. So here, if you have to replace R with Rmax, then you can replace dR by dθ with zero. I am pretty sure you agree with me. We just sidestepped that problem and now we have a very clear equation. Let’s see what that equation is. So this is one term that we need and this is another term that we need. So we have to bring them together and write our answer. So what will that look like here? Rsec2θ, yeah, minus this term only right? g by 2v2R2. So you have a 2secθsec2θtanθ. So you have 2secθtanθ, where you can cancel out the 2, so we will remove them. So this is what your expression will look like. This whole thing equals zero. But the R here is not just some range but Rmax. I am not going to continue writing Rmax throughout so I am writing it over here, because it is going to be long to write, right? So finally you know the R that we find is the maximum range. The maximum range satisfies this condition, along with θ. So let’s go and find out what that’s going to be.
So we will copy this down over here. So this is our relationship. One thing you observes is R comes out, yeah, first itself. 1R comes out, sec2θ minus g by v2 R into sec2θtanθ equal to zero. Yeah, actually we notice sec2θ comes out as well. So if I write sec2θ outside, yeah, I can remove this as well and write 1 and 1 in its place. I have just taken sec2θR outside. So this means that R must be zero because sec2θ is clearly not zero. Yeah. Why? Yeah, think about it. You know that sec2θ is not going to be zero but the range can be zero. But that corresponds to the minimum range. So we are cancelling this out and continuing with this equation. What we have arrived at is this corresponds to the minimum range. And this makes sense right? Because when you equate the differential, or we equate the derivative equal to zero, you don’t just get the maximum, you also get the minimum. Mark doesn’t know whether you are looking for the maximum or the minimum, it just gives you the points at which the curve has zero derivative. Which can either be a minimum or be a maximum. Already you’ve got the minimum value, it’s zero. You know at what angle that will happen. Tell me? 90 degrees. Right? Throw it up, it will go up and fall down – the range would have been zero. That’s the easy part of the problem. This is done.
What we need is a maximum range. So, let’s continue with this part of our equation because we know this is also equal to zero. So we do that then what do we get 1 minus gR by v2tanθ equal zero. Which will give us, what? tanθ. If I take it to the other side, it will be equal to v2 by gR, where R is your maximum range. When your range is maximized, your tanθ and your max range will have this relationship. Watch that this does not have s in it. So this is a very simple explanation. Now what is your next step? So sometimes when you are doing a lot of math, it’s a little difficult to keep track of what’s the final thing we are after. It’s almost like life. The nitty-gritty details of life takes you away from the final goal. And I remember my math teacher, who is one of the best human beings I have ever known. Very inspiring man. So when he taught me, he was really old, like around 70 or so but most of his metaphors were football metaphor. And I remember him telling me that it’s important to, when you’re playing football, it’s important to look at the ball but it’s also important to keep an eye on the goal. And either of these can’t be missed, keeping that balance is really crucial. So I think math kind of trains us to do that. So we did a lot of things and we finally came here, but now we have to answer the question why did we come here? We came here because we wanted a relationship between Rmax and θ, so that we can substitute in our first equation and get whichever we want. In this case we want the θ at which we should launch so that we get maximum range. We don’t want the maximum range itself, right? Because we know 45 may not be the answer. We’re trying to find out what that angle will be. So will this achieve that? Yes it will, because you can rewrite this equation as, this right? You can write this as Rmax equals v2 by gtanθ. Do you agree? So I have just taken the Rmax here aside and tanθ below. So let me copy that down and let’s put it into our first equation. Which equation? The equation which we wrote after eliminating t.
So great. We have our two equations. We are going to plug the value or Rmax into this so that we get our value for θ. Because when you write max over here, the θ that you get will be the θ that corresponds to the maximum range and which is what we want. So let’s do that now. So we will get v2 by gtanθ into tanθ minus g by 2v2R2 right so v4 by g2tan2θ. I am trying to be as careful as I can. Into sec2θ equals s. So good amount of simplifying that we can do, yeah? So we have tanθ getting canceled out. We have v2 by g over here and you have, similar v2 by g over there by 2g, that is, and sec2 and tan2, what can you do with that? These two will cancel out to give you sin2θ right. Why? sec2 is 1 by cos2 and tan2 is sin2 by cos2. Those two will cancel so you’ll get this. It’s little easy to verify. So if we have v2 by g minus v2 by 2g. Am I counting all terms? 2, g and sin2, one sin2 I have to count. Right? Equals s. I am getting closer and closer to my answer right. This θ is what I want. You always have to keep track of what we want. This θ is right now stuck in the middle of constants. Every single other thing is a constant. Which means I am terrible close to my answer. Yeah. So all I have to do now is rearrange this equation to get whatever answer it is that I need. So what will I do? I will take my v2 by 2g as common. So v2 by 2g so that I will multiply with 2 inside. So that will work. There will be 2 minus 1 by sin2θ and that equals s. Yeah. I am trying to isolate. What I am trying, why am I doing this step? Why am I taking this common all that? I need to find a way to isolate my sin2θ from everything else. Theoretically I have my answer, I could have substituted numbers and got it. But if I want an answer in terms of these variables, then I have to do the hard work of splitting them. Looks like a bit of hard work. Let’s see. Let’s copy it down to the next page and let’s see what we can do.
So now what can I do? If you look at this now, let’s see, I can take this constant to the other side, yeah? Which means it will give me some more leeway but before I do that let me try and simplify it this term over here. Yeah. It is a trial and error game okay. By the way if somebody is doing it straight, it just means you have done it sometime before. So we are trying out this and that to see how I can isolate this term. That’s pretty much what we are doing here. So you will get v2 by 2g, yeah, into 2sin2θ minus 1 by sin2θ, equals s. Good till here. Let’s do one thing. I’ll tell you why I’m doing this okay. I am going to put a minus here and multiply this equation also with a minus. Why? I feel one minus sin2 is a more useful quantity than sin2 minus 1. Yeah. I will tell you why because I know that 1 – sin2 is cos2. So basically what I am trying to do this trigonometric quantities is trying to use some identity. Because I don’t know the value of θ. If I know the value of θ, then I am done, I wouldn’t do this. Some identity I want to use should work for all θ and simplify my expression. So that’s what I am doing here. So this means it will be 1 – 2sin2θ. Do you agree? That is equal to 1 – sin2θ – sin2θ. Wow. This looks like I am going somewhere because now this will become, this part right, will become what – cos2θ. Right. 1 – sin2θ is cos2θ. So if I replace that now, I’ll have cos2θ – sin2θ by sin2θ equal to minus s. So now what can I do? Don’t ask me why I am doing this okay. I am doing this because I don’t want to solve for both cos and sin. I want to solve for only one variable, so I am dividing the whole thing by cos2θ. So what will happen if I do that? This by cos2θ, this by cos2θ, this by cos2θ. Now good question to ask here is why will it strike me to do this. By observing a pattern only. I don’t have any other answer. If you observe this pattern you know you can get by dividing by cos2 θ you know you can get something that looks like this, v2 by 2g into 1 minus tan2 θ by tan2 θ equals minus s. Now you might ask what progress have we made from here to here. You’ve brought it down to a form where there is only one thing that contains θ. And that’s very close to victory okay. Because here you’ve two. How we do it? Here you have only one, you’ve got closer. We’ve used clever mathematics to do this. You should feel proud. Now what do we do? Rearrange and get the answer right?
For me as of now my tan2 θ is just some variable right. I don’t care what it is. I will say, let my tan2 θ be equal to some t. Right. And let’s rewrite the same equation. Now this is no complex than the previous one except that it looks cleaner for our heads right. It’s like the psychological simplicity over here. So from here where you can go is rearrange the equations so you’ll get minus 2gs by v2 by taking this to the other side and this also to the other side equals 1 minus t by t. Yes. So let me again I want to, I want some psychological simplicity over here so let me call this entire part right. I know the whole thing is a constant. Minus 2gs, let me call that whole thing equal to some constant K. Then what will my equation look like? 1 minus t by t equals some constant K where the K is equal to this entire, with the minus sign, equal to this. I am doing all this to minimize my own error okay. So here, then you will get 1 minus t equals Kt or t is what you want right, so t equals 1 by 1 plus K. Yeah. Now you can substitute back whatever you took. This kind of made it simpler, right. So we substitute back what you had for, so t will be tan2θ, yeah. That will be equal to 1 by 1 plus, K is minus 2gs by v2, so we will replace this with minus. So we will have minus 2gs by v2. So if we write it down again over here you’ll get tan2 θ equals v2 by v2 minus 2gs. Very, very close. Now we just have to take a root and that’s okay. tanθ equals, or I will jump a step here, I’ll write θ equals tan-1 I feel really proud okay, that I have come this far and I feel even more proud that you have come this far with me. But we’ll analyze our results. v by root over v2 minus 2gs. What does this tell us? The angle at which you must throw, if you have to throw, such that you catch it some vertical displacement s. What is the obvious step here? After doing all this hard work. Verify. How verify? We know one answer no? When our s is zero we know our answer must come out to be 45. In other words tanθ must come out to be 1. Let’s hope, let’s pray that it’s coming. You know, so you have, s will be zero, right in that case? So let’s do our special cases. So for s equal zero we expect 45, but let’s see what happens. This will get cancelled, yeah. So what will you have? v by root over v2, and you are taking positive values here so you’ll get tanθ equal to 1. A sigh of relief. Doesn’t prove anything but we are very, very close to the, it’s very unlikely that we are wrong here, right? Now I did verify okay this is correct. So don’t worry about it. Don’t be worried about how unsure I sound. This is right, what we have done here. We have not made any mistake throughout. So your θ for getting a range max equals this.
So having made this journey let’s look at the implication. So your tanθ comes out to be this value and we already saw when s equal zero your θ came out to be 45 as expected. Now by the most important observe there is an s over here in this expression which means it does matter. So let’s see, if you want to throw from a height, right, will your, what can you say about your angle. You can sure say if you know this to be s right, then if the value is s, then for you something will be minus s. Yeah. Let’s say you are throwing from some height H, then your s value will be –H right? Because it’s downwards. So what will your equation look like? Your tanθ will be equal to root over v2 by v2 plus 2g into H. Yeah. What is to observe here is that this value is always going to be less than 1, which means the angle will be less than 45 for sure. So the intuition for this is, what’s happening is this: if I have something thrown at 45, it’s going like that and landing here. Something thrown at some other angle, we don’t know what that angle will be. Not all angles below will do better than 45, but some angle will. What it will do is there it will be a loser till the horizontal. Yeah. But after that it will beat our competition, which is 45 degrees. So if this is 45, the blue is 45 and the green is the one that follows this equation. Then we know what’s happening. Till here the clear winner is 45, but after that it will be the angle less than 45. There is some intuition here as well. If some angle greater than 45 had been thrown, we can for sure show that it cannot beat because we know it will be before 45 till horizontal, that is 100% sure, 45 will be the winner. But after that this has a steeper velocity right. Yeah. Which means it has lesser time and a lesser horizontal velocity how will it beat? Something has to be better. What is better for our 30 degree case, or I am saying 30 degree, some angle less than 45. In fact 30 is a very wrong thing to say. For that case, the horizontal velocity component is larger and it over here is more along the horizontal that it is, it also gives it some more time so there is a chance that it ends up crossing over after the horizontal has been reached. And that’s what we have seen here.
If the question had been something like you have to clear a wall of some height H how far away can this wall of height H be so that you can still clear it? If that’s the question then your height would have been a positive number, so the same θ would have been, let me use a different color for this, would have been tanθ equals root over v2 by v2 plus, yeah, but all this expression it’s a minus right? So it would have been a minus and in this case your s is just plus H. There is no minus to cancel out the other ones so you will have v2 minus 2gH. Now what this shows is the angle for a question like this. If you were ever asked there is a wall of height some 10 meters. You are to throw a ball such that it will go to the other side of the wall. How far can this wall be, given your speed? Then the angle at which you should throw, such that it gets at maximum range will be greater than 45 degrees. And if you are throwing from a cliff it will be less than 45 degrees. And if you are throwing and catching at the same height you know what it is. It will be exactly equal to 45 degrees.
So hope you have enjoyed this journey. It was slightly long, involved some calculus, but this is general case. It’s a long problem but it gives you some intuition about how projectiles behave when they are thrown not on a plane.