So, because you already know about projectiles, I know you want more challenging questions. Let me begin with one. Let’s say I threw three projectiles: one like that, one like this and one like this, 30, 45 and 60 degrees. Now the question is which of them will last longest in the air? So you know how long the ball will stay in the air depends upon? That’s right. The vertical component, right? Does it have anything to do with the horizontal? It doesn’t. Then, as you can see, all the three bodies have the same vertical component. So all of them should remain in the air for the same time. That’s our intuition. Let’s see if it works. Now did you notice something there? What about the height to which the balls went? Yeah, even the height that they reached was the same, which means that the horizontal component has no say, even in the height. So neither in the height, nor in the time of flight, the horizontal component have any say at all. So now let’s flip the diagram, so that all the three balls have the same horizontal component, but different vertical components. Then if I ask you, which of them will land first, what will your answer be? Now clearly the ball with the least vertical component will have the least amount of time in the air, therefore it will land first. Must be simple here. Then the next ball and then the ball with the highest vertical component. All this must also be true for the heights right? Because these two are very, very closely connected. But the interesting thing here is that the moment you launch these three balls, throughout their journey, until one of them lands, all three of them will be in the exact same vertical line. Which means if one of them, the bottom most let’s say, looks up, he will be able to see the other two right in the same line. Now once the bottom most one lands, because that will land first, the other two will continue to follow this rule. With all this, if I ask you, of the two motions, the vertical and the horizontal, who’s the boss? What will your answer be? What does the time of flight depend upon? Only the vertical component. What about the maximum height reached? Only the vertical component. What about the range though? The horizontal component? And the vertical, right? Because the range depends on the horizontal component and time, and the time depends upon the vertical component. So who really is the boss? The vertical component. It calls all the shots. It says how long the ball will be in the air, how high it will go. The horizontal just determines, given the time, how far the ball will land, on the x-axis.
Now, in the above question, where we threw three balls at 30, 45 and 60, if I asked the question, which of these balls will travel the farthest, while they land. In other words, which one of them has the maximum range, what will your answer be? Now, I know some of you were tempted to say 45 degrees right? Maximum range means 45 degrees, without any question. There is one other problem with remembering formulae or statements without any context, right? The forte of modern journalism. Take statements out of context, use it everywhere. But let’s watch what happens. Now what you saw, should not surprise you because all the three balls have the same horizontal component, which means that for a given second all of them will travel the same distance. So what’s going to determine the range? The time spent in the air, and that’s determined by whom? The vertical component. So the ball with the largest vertical component will have the maximum range over here and that’s the ball thrown at 60 degrees. But now you will say, “Sir, but maximum range somewhere I remember from is 45 only.” Where does that come from? So we will look at that now.
So the beginning, we kept the vertical component fixed, and then asked some questions. Then we fixed the horizontal and asked some questions. Now let’s fix the result end of the two, in other words, the speed. And then ask the questions at what height and range. So we have three balls thrown like this, but the speed is the same but the angles are different. Now we ask which will reach the maximum height. That shouldn’t be a problem. The one with the largest vertical component, which is the ball thrown at 60 degrees. But what if I ask you which will have the maximum range? So I will look at this and just by looking I am not able to make an inference. Because here I am throwing and catching at the same height. Maybe I can use the result that I have, for range, which is the time multiplied by the horizontal velocity. So, u cosθ, the horizontal velocity, multiplied by the time which is 2u sinθ by g. And when I look at this expression something stands out, right? Which is that cos and sin are symmetric. In other words I replace one with the other, it won’t make a difference to the result, which is the range. And when does cosθ get replaced by sinθ? When we replace θ with (90 – θ). So the point I am trying to make here is, looking at this expression of range, just mathematically, it suggests that the range will be the same if we replace θ with (90 – θ). In other words, 2 balls thrown at angles θ and (90 – θ) will have the same range. That’s what this is suggesting. But that’s counterintuitive, right, that how can two balls which are thrown at two different angles have the same range? It doesn’t seem right because they are spending, or they are spending two different times in the air. Okay, so this makes sense. One of them has a large horizontal component but spends a short amount of time because of a small vertical component and the other has a small horizontal component so it doesn’t travel too much in a second, but travels longer in the air because it has a large vertical component. So in one sense the time and the horizontal component are balancing each other out. So it does make some intuitive sense. Let’s watch how it looks like. Wow! How beautiful is that? You can see that complimentary angles have the same range as long as you are throwing and catching at the same height. Which means in our original question, 30 degrees and 60 would have had the same range. So we have learnt a lot already, but still we don’t know the answer to one question right? Which of the three will have the maximum range? To calculate the maximum range let’s look at the expression again. But now, let’s do the same rearranging we did back in the previous video where we told you, right? The reason for rearranging will become obvious now and it’s going to become obvious here. So how are we going to rearrange? 2 times sinθcosθ is sin2θ. So your expression is going to look like this. u2sin2θ by g. Your u is a constant, g is a constant, 2 is definitely a constant. So you have sin2θ over there. So this expression is going to get maximized when the value of sin is going to get maximized. What’s the maximum value sin can take? Yeah. One right? It can never go beyond that. So the best we can hope for over here is u2 by g. So for that sin2θ must be 1. When is sin2θ one? When that expression inside, 2θ, is 90 degrees, right? sin 90 is 1. So 2θ is 90, becoming obvious that θ must be 45. I know some of you are sighing a huge sigh of relief. Some familiarity at last, right? Maximum range of 45 degrees. Now you know how it comes. It comes from maximizing the expression for range. Now I want you to watch something closely here. Right? Is there anywhere we could have inferred anything about the question asked to where throwing from a foot. Have you ever spoken about it? The best answer to this question should have been, “I don’t know.” Because we never discussed that case. Till now what have we been talking about? All our results were for a very special case, right? Throwing and catching at the same height. So there is suddenly a question about something else then the answer should be I don’t know, until we go ahead and solve it. So, let’s be very sure about what inferences we have made, and what we have not. What we have made in this video is that if I throw at a constant speed where I can only change my angle and if I throw and catch at the same height then if I throw at zero degrees, the range will be zero. As I keep increasing my angle, my range will keep increasing until the angle is 45 where I will reach my maximum range. And then, as I increase my angle even more, the range will continue to reduce such that the complementary angle keep having the same range and finally it will come back to zero. Therefore, when thrown and caught at the same height the maximum range is 45, right? It’s not true in general. But I know that you are not satisfied. I know you want also answer questions about throwing from a height and landing below or throwing from below and landing at a height. I know you also want to answer the questions about what’s the maximum range in these cases. Because you know it need not be 45. Now, these are more challenging and therefore more interesting questions, but the good news is you already know how to solve them. So if you are good and ready, let’s dive in, or let’s say, drive in.