I think most us love trying out new adventures or watching a circus act. The human cannonball is an old circus act that takes place by a specially designed cannon to shoot out a human. The human then lands on a safety net which is placed to provide a safe landing for the human. Well, this is the important question “Where must the safety net be placed?” And this question can be answered by classical mechanics.
Before this act takes place, there are question questions that the human will have, for example, They would want to know how high the ceiling must be if you are doing it indoors or how long will he be flying, waiting to land on the net?
Let us see how we can answer these questions.
What’s in our control? The angle at which you can throw and the speed which you can throw, u and θ. Let’s solve how long would it take to stay in the air.
Answer: If I had thrown the ball up, how long will it take to reach its highest point? u/g right? Then for it to come back again, it will be 2u/g. So if I had thrown a ball up, it takes 2u/g seconds to go up and come down. Now what is this whole projectile motion viewed from this point of view, it’s just a ball thrown up at u sinθ. So you have 2u/g, you just have to replace u with u sinθ and you get the answer for how long this person will stay in the air.
For the question of how high the ceiling must be?
From the front view, because that’s the view in which you see something going up and coming down. If we throw a ball up at u, how high will it go? u2/2g. it will take you exactly 5 seconds to derive it using v2 – u2 = 2as. So if you have u2/2g, that’s if the ball is thrown up at u, but here its u sinθ, so just replace, again, u with u sinθ. You get (u sinθ)2/2g. If you want to write it in another way, you can write (u2 sin2θ)/2g.
Here, we have seen few questions how the safety net can be placed at the right place. Watch the video given to answer all your questions and have a complete understanding of the concept taking place in a human cannonball.
Faculty: Have you heard of the human cannonball? It’s this old circus act in which a human being is launched from a specially designed cannon and is caught far away with a net, or a body of water. Now the question of where do we place this net is dealt with by classical mechanics and that’s where we step in. Now let’s say you are a hero, and you want to take this daring act. Then, what’s the first question you want an answer for? Where must the safety net be placed? And there is no question of trial and error here right? You won’t accept if I come and tell you, “Hey, let’s keep it here. You jump once and let’s see what happens.” That’s not something you are going to accept. Now also, you will want to know how high the ceiling must be if you are doing it indoors right. Otherwise you will go hit the ceiling and then it doesn’t matter where you place the net.
So you want to know that as well. And if you are a pessimist and if you decide that these are the last few moments of your life, you will want to know how many such moments exist. In other words, how long will you be there, flying, waiting for yourself to land on the net? Now how do we answer these questions? The first thing I want you to notice is are these new questions? We already answered them right, before? We took a very mathematical approach, where we solved projectile motion. So we found the value of time at which the y becomes zero, found the value of x over y becomes zero, found the value of y when the velocity becomes zero. But all this was very mathematical right? So in this particular video what we will do is we will take a much more intuitive approach to the very same questions. What’s in our control? The angle at which you can throw and the speed which you can throw, u and θ. Now these questions take only a few minutes to answer.
Once you understand this it will take you less than a minute to derive the questions to all these answers. And we will prove that to you at the end of this video.
Let’s solve the pessimist question first, which is how long do you have to stay in the air, the time. Now when I was thinking about what would be the easiest way to explain this or visualize this right? Because I know we have to visualize it as two separate components. But how do we do this? I was talking to Ram, who’s famous from:
Ram: Guy’s I have taken so much long time to make this adaptive flow, to put on the right cushion in the right place. From the simplest to the toughest. Please go and solve. Please.
Faculty: Yeah, the one who was begging about the adaptive flow right? Where you can solve problems. Now we were discussing and he was asking why are we so stuck up with this side view of a projectile. What other way can we look at the same thing? What would happen if instead of this view we were to fly up and look at the same ball being thrown from the top? What would that look like? It would look like something is moving in a straight line. Right? Just one dimensional motion and at a constant speed of the horizontal component which in this case is going to be u cosθ. Interesting. Now is there another way we can look at it? What if we go to the front view, in other words, the view in which it looks like a person throwing right at you, and then watch the same thing happening. Again it’s going to look like the body is moving in just one line. It’s going up and coming down. Right? Except that the speed at which it starts will be u sinθ, the vertical component. Now to answer the question how long will you be in the air, which of these two points of view do you think is more useful?
If you are to look at this motion from the front view, you will notice that it’s as good as a ball being thrown up and coming back down again. And that is familiar. Isn’t it? So, let’s bring that over here. If you were to notice, if I had thrown the ball up at you, how long will it take to reach its highest point? u/g right? If you are curious why, every second the velocity reduces by g, for it to become zero it takes u/g seconds. It’s easier to do. Then for it to come back again, it will be 2u/g. So if I had thrown a ball up, it takes 2u/g seconds to go up and come down. Now what is this whole projectile motion viewed from this point of view? It’s just a ball thrown up at u sinθ. So you have 2u/g, you just have to replace u with u sinθ and you get the answer for how long this person will stay in the air. A change in point of view can make our problem really simple, can’t it? Now think about this again. For the question of how the ceiling must be, which point of view do you think is useful?
Again the front view right? Because that’s the view in which you see something going up and coming down. So again in our 1-d motion we knew this right? If we throw a ball up at u, how high will it go? u2/2g. it will take you exactly 5 seconds to derive it using v2 – u2 = 2as, if you want to. So if you have u2/2g, that’s if the ball is thrown up at u, but what’s the ball thrown up over here? u sinθ, so just replace, again, u with u sinθ. You get (u sinθ)2/2g. If you want to write it in another way, you can write (u2 sin2θ)/2g.
Which brings us to the last and the most important question right? Where should you keep the net? Now for this question, which view is important? Yeah? You’ll understand that both views are important. But if you take the top view now, so if you tilt it and if you are able to see the motion, uniform motion along the line, all you’re trying to find out is how far on this line you should place the net. Right? You know the speed at which the motion is happening. So what’s the only variable you need? Time.
The pessimist question was answered the beginning a little cleverly, so that we can use it over here. So what do you have there? Some distance that you must keep the net at. You need that distance. We have the velocity, we have the time. This is an eighth standard question okay. Find the distance given the velocity and the time because it’s uniform motion. So what’s your velocity? u cosθ. What’s your time? You already have it right? u sinθ/g, twice of that, so 2u sinθ/g for it to go up and come down. So you have the time, you have the velocity. Your answer is going to be just the product of this. Now we can arrange this, okay, to make it look different. And you’ll notice why we are doing that in the next video. So we will do that over here. What are you getting over here? Yeah, u cosθ multiplied by (2u sinθ/g). Yeah? So two times u cosθ and u sinθ is sin2θ. Are you asking me why? Okay fine.
If you’re asking me why, I will tell you this. There is a small identity in trigonometry which is sin (A+B) = sinA cosB + cosA sinB. Over here you put A and B as θ and you get sin(θ + θ), which is sin2θ, equals sinθ cosθ + cosθ sinθ which is 2 sinθ cosθ. Some of you might also ask why is sin (A+B) = sinA cosB + cosA sinB. We will not go into that derivation here because I don’t know the derivation. Also, because it may be a little irrelevant, isn’t it? So in the final form, the question of where to place the net, looks like this. u2sin2θ/g.
Now these three answers have very serious names okay. The question, how long will you be in the air is called the time of flight of a projectile. And how high will it go is called the maximum height that a projectile reaches. And how far you should place the net is called the range of a projectile. Now, don’t remember these results. Why? Because remembering or recollecting these formulae takes longer than deriving them.
You want to see how? Let’s see on the time of flight. You know a ball thrown up at u takes u/g to go up, u/g to come down, so overall 2u/g. Therefore you’ll replace u with u sinθ, (2u sinθ/g). You’re done right? On the maximum height, you throw a ball at u, you know it goes to a height of u2/2g. So let’s replace u with u sinθ over here. (u2 sin2θ/2g). Done. On the range, you know the velocity is u cosθ initially. You want the time, time is already over here. So u cosθ into (2u sinθ/g). You can rearrange it, you’ll get u2sin2θ/g and you are done.
That’s all it takes to derive all these problems. Doesn’t it? And that’s not even the only reason to not remember these. There is a more important reason and that is that these formulae for the range, the time of flight and all that are for a special case. Right? We’re throwing and catching at the same height, right? So when you call this the range of a projectile, you’re not mentioning, for the special case of a projectile that was thrown and caught at the same height. And in most good exams you will not have a problem which expects you to just plug in a well known formula and get the answer. There will be some change and remembering this formula will show you, might actually constrain the way in which you think.
Instead of taking a simple approach, you might actually end up taking a more complex approach because you remember these formulae. Even if you do want to internalize these results, the best way to do that might be to solve problems and whenever you need these results, just derive them. And eventually you will reach a point where you don’t have to derive them at all. Oh yeah, except for those special cases where the derivations are really, really long okay? Calculus might be a good example. There remembering might help. Now, we wanted to badly shoot a footage of the human cannonball going there and being caught in the net. Unfortunately, nobody volunteered. So we had to do something on a much smaller scale. So here it is for you. So if you did know the angle at which the car left the track and the speed at which it left you can answer the question where should you place the track so that you can exactly connect the car again. And for a change you can try this at home.