### Jumping off Cliffs

So till now what have we seen? Projectile being thrown and caught at the same height. Just before the previous video we saw how in general it can be caught in other ways as well, right? So how many projectile problems are there? Exactly one. Even though it might be dressed to look differently. So what all can you do with a projectile? You can throw, you can catch at the same height. That we already did, so that you can see. The others are that you can throw and catch at a higher height or at a lower height and that’s pretty much it. Right. That covers all your cases. So now what we will do is we will begin with one of the questions that will seemingly do the first case, which is throw here and catch somewhere over there. Let’s solve a problem like that. Now you will notice how easy this is. Let’s say I have my familiar Angry Birds scenario. So I am launching an angry bird, yeah? So you can see over here an angry bird is being thrown with a velocity of 10 root 2 meters per second at an angle of 45. I want you to find out how much time Mr. Piggy has to live? It’s a pretty simple question right? So this body is passing through the point, let’s say 10, 15 and then you are done. The y-coordinate of the point is 15 and the x-coordinate is 10. And with that you are good. So, what’s the first step of any problem like this? Take components of your velocity and forget the original velocity. So let’s do that. Your vertical component will be 10, your horizontal will be 10. Yeah, we chose our initial velocities to make it so that the components come out easy. Now is not the time to make the math difficult, right? Now we will get the concept and then we can make the math look very ugly 10 root 2 by 3, something can happen. Not here. So we have this at two now then, to answer the question how long you actually don’t need any time at all because you are travelling with a horizontal velocity of 10 m/s that never changes. There is no acceleration. The only acceleration is g downwards. You are travelling a distance of 10 m. So how long will it take? 1 second. You are done. It’s like the shortest problem you have ever solved, right? Even though it’s not in the plane, it looks difficult. Yes. Your answer is correct. You’ve just made one assumption. What is that? You’ve assumed that this angry bird will hit that pig. It’s not confirmed right? If it is told when will it hit the pig was the question. But will it even hit the pig? Did you think of that? Because if you didn’t think about that, this answer is correct. Right? It’s starting at 10, travels a distance of 10. Uniform motion 10 by 10. The answer is 1 second. But let’s just see. At this time once again what the height of this angry bird will be. Yeah. Let’s see what the answer to that is. So if you go back one step over here the initial vertical velocity right, slated for the Y-equation. Let’s call it the Y-equation. The equation for the y-coordinate. What is the displacement of the time t? Or what we want it to be? We want it to be 15. We will check what it is and then see if our answers match. So your u sin theta the vertical velocity is 10. You’ve already calculated it. You don’t care about this anymore. Yeah, this is not required, right, in your entire figure. Let’s just remove this. So we have 10 into time t which is 1 second minus half into g is 10, your t square which 1 squared again equals your Y after time 1 second. So clearly the answer is going to be less than 10. So, it certainly is not going to be 15. What’s the answer? 15 minus 10, 5 meters. Too bad. So what’s the real answer to the question? So actually our projectile is going to look something like this. Hitting it somewhere here. This will be 5 approximately. So who will be laughing at the end? Our man here because how long does he actually have to live? Infinitely long because the ball doesn’t hit. Just pointing out a way of thinking as well but as a problem this is very, very simple, right.

So let’s look at one common problem that is asked. It’s a question that comes very, very commonly. Let’s setup the problem. So the common problem that you would see is that a ball or anything at all, say an arrow or a missile is being launched from a cliff from some given height, yeah? And you have to find out what range it will cover along the horizontal and finally land somewhere over here, right? So, it’s kind of like asking you are throwing something from a table. Yeah? How far it will go and land on the ground? That’s pretty much the question. Now when you are doing this – how do you answer this question? A large number of students who I see in class, even though I warned them beforehand to notice, have a tendency to cut this diagram somewhere along the horizontal as if this point in the motion is a very special point. Nothing is changing at that point. But still we have a tendency, a let’s call it a temptation to cut it over here. You know why? We have that temptation, or most of my students have and I ask them “why do you do this?” and they say I know the formula for the range till here. Range formula. What is the range formula? u square sin 2 theta by g that I know. Till here. So what they do is they calculate till here and then they will be sitting like this and asking ok from here I know the range. What do I do from there? Yeah? It’s not very, it doesn’t sound like a very smart thing to do. I will tell you why. Because anyway you have to solve this remaining part of the problem that’s as complex as solving this problem only. So we might as well solve this problem directly. Yeah. Except that what they’ll want, what you might want to do is, you know that this range has to be at least larger than this range so you can eliminate options that are larger than, or sorry, smaller than this particular range. Something you can do but if I make options I will not make them in a way that you can eliminate them so easily. So you can’t correct here. Another common thing that I notice of course is where you cut the diagram over here at the top most point. Till there you calculate the time, find out this range and then add it to this range, or whatever it is, right. So my claim is all of these are not effective ways of solving the problem. Why? Because once the ball was launched it did not care about anything else. Its gravity was always downwards, it was uniformly accelerating. It just was caught somewhere below. Who cares? Yeah. The equation will take care of all that because you will not think of a projectile motion problem this way right. How do you think of projectiles? You think of them as a ball that is thrown up which goes like this and comes down. A 1-d motion problem, which also happens to cover something along the x. It’s as good as the ball was dropped, gravity suddenly stopped over here and then the ball went flying like this and stood somewhere there. Yeah. Who cares? Do you care whether the actual path looked like that and all? Do you care about all that? You don’t. That’s the whole beauty of taking components. So to take this approach what do we have to do? First break down our motion into our two components which in this case will be very easy to do because the angles have been chosen that way. So your 40 root 2 and a 45 degree angle which basically just means that it’s 40 along this side, 40 along this side. Sin and cos 45 are 1 by root 2. So 40 m/s, 40 m/s. We are good. This height is 45 meters. So we bring it back over here. So we have 45 over here. You don’t care about the horizontal motion. Why? Who calls the shots? The vertical. This ball moving is the important part of the motion. You can think of it as a ball thrown like this, a projectile is basically a ball thrown like this is basically just a ball thrown up which also happens to move something horizontally. I am saying happens to because the time that is allowed for this horizontal motion is determined completely by the vertical. So given that, all you have to do now is calculate this time. What you need is the time. That’s the bridge that takes you from the vertical to the horizontal. Why? If you find out how long the ball remains in the air then multiply that with the horizontal velocity, you are done. That will be the distance covered. That’s exactly the same thing you did for horizontal range also ok. But the reason why we attempted to cut the diagram here is that formula you remembered. So sometimes remembering can constrain us. We try to stuff another problem even though it can’t be stuffed into a formula and that can actually slow us down instead of speeding us up. I am not saying remembering formula is bad all the time. Yeah. It’s good to know what happens to arrive at the formula so that you can use it in other places. So you have that over here. So you have 40 up, g is downwards. It’s 10 usually. Yeah. If it’s mentioned its 9.8 then take 9.8 okay in general. We always take 10 in our case. Yeah. So 45, 40, 10. Do you need anything else? Nothing right? So write the vertical equation. Have I drawn a projectile problem diagram here? No, I have drawn a 1-d motion diagram problem. Because it is a 1-d motion problem. So I have 40 into my time t minus half into g, which is 10, into t square. u t minus half g t square where u is going to be positive 40 because my convention is up positive and right positive. Right? So we have this. It’s going to be equal to what? This is one place where you might have to be a little careful. It’s not equal to 45. The displacement is not 45. What is the displacement? It’s minus 45. Because you are taking up as positive then this displacement is in the negative direction as in downward direction. We are taking displacement as downwards is negative. So this is your equation. It has only one variable t. Solve for it you will get the answer. Let’s do it. So it’s going to be. You take it to the other side, it’s going to be 5 t square minus 40 t minus 45 equals zero and if you simplify it t square minus 8 t – 9 equals zero. Now this equation happens to be factorable right you can see here. t minus 9 and t plus 1 will be your solutions, will be the factors for this equation. Yeah. t minus 9 and t plus 1 will give it, which means the solutions are t equal to plus 9 and t equal to minus 1. Now you know what the minus 1 means here so I am not going into that. Minus 1 over here will imply the time taken for a ball launched here to reach here with the velocity of 40 m/s. If I had thrown the ball up here with some velocity, we can calculate that. For it to reach here at 40 it would have taken exactly 1 second. What is the more important answer for us is this ball thrown at 40 takes 9 seconds to go up and come down and with that your answer is solved. Right. Almost solved. Nothing else. What do you have now? You know the ball gets the time of 9 seconds to travel this horizontal distance. With what speed will it travel? This horizontal ball. 40 m/s. Right. Because we calculated our horizontal velocity to be 40 m/s. So you come here. 40 m/s for 9 seconds. Yeah? 40 m/s into 9 s equals, yeah, 40 into 9 is 360, seconds, seconds will cancel, meters. 360 meters is what it will travel along the horizontal and that’s going to be the answer to the question. How far the ball would have gone? But you don’t have this, you don’t need this picture right. Like we told you, you only need to calculate it as going up and coming down separately. This motion separately.

So let’s think about one more problem, which is also interesting to listen to. I again notice that you will be applying the same fundamental principle of writing the equation for the vertical, finding out time and then going to the horizontal. Time will be the bridge between your vertical and your horizontal. Let’s setup the problem. So that’s our problem setup where a batsman hits a ball. Let’s say Chris Gayle and the ball’s speeding. It’s in the air though and as a fielder, unrealistically tall fielder, 3 meter tall fielder which is virtually impossible. 3 meters because he is able to stretch his hand out, right, that final distance is 3 meters. A really tall guy. Now he wants to catch the ball. Right? Now he is 30 meters away. What must be the minimum speed he has to run to be able to catch the ball? Fair thing? Now we need some more information right? What is that? What’s the speed at which the ball was hit? The ball is hit too fast before you run it will just go away. So it’s 10 root 2 m/s. I don’t think that’s too unrealistic. So our problem is not too unrealistic. Hit at an angle of 45. So the batsman is being extremely mathematical. He knows 45 degrees will give him maximum range, highest probability of a 6. So he is hitting it 45 degrees. Now this fielder over here is the one who has to think and realize how fast or what must be his minimum speed to run. So to do this what do you have to think about? Let’s go into the thought process of solving this. So we have a fielder whose basic catching height is 3. So your problem solving ability is to break down this problem into its bare minimum which is basically when will the y-coordinate of the ball be 3. Only then he will catch. Yeah. He will not catch the ball when it’s down or he will not jump up and catch it. He will catch the ball exactly when its 3. That’s the assumption in the question. Yeah. Because if he catch it higher, if he is really, really tall right. He will just grow tall here, put his hand here and catch wherever he is standing. He doesn’t have to run. So I am telling all this because these are all implied in the question. And many time a question implies these kind of common sense things. So he is constrained by his height even though he is so tall the ball is going above him. He can jump up right here and catch the ball. Yeah, like we showed. But clearly it’s not possible. 3 meters is the maximum we can go to even if he jumps up let’s say. Then he has to run some distance till that ball comes to at least 3 meters, then only he can catch it. Yes. So let’s ask that question. When will that ball be at a height of 3 meters? Right. So you have an initial velocity of, that’s right, 10 root 2. Will you ever use at an angle of 45, but you will never ever use that directly. What will you do? You will break it down so you have your broken down components. 10 over here and 10 over here. Then you don’t need these. Great. So 10 vertical 10 horizontal. Your problem boils down to something very simple as a first step. What’s the first step? When will the ball be at a height of 3 meters? Only then he can catch. If the ball is at a height of 3 meters only after the boundary then there is no point. But here let’s find out the answer. So initial velocity in the vertical direction. uy equals 10. Acceleration equals minus 10. Right? Now your time is what we want, but what is given? The s, or the height at which he will catch. Yeah, that is given to be 3 meters. Correct. So now if you write the equation and solve this, you will realize how repetitive this is right. Actually almost every problem we are doing this only. Even though they can look like different problems. So just do that now, you will have, use a different color. You will have uy into t minus half g t square equals 3. Plus 3 in this case. Yeah let’s substitute the other values also. So it will be 10 into time minus 5, because u by 2 will be 5, t square equals 3. Yeah. So now if we simply this what will you get? So you get 5 t square minus 10 t. So you have 5 t square on one side, minus 10 t on the other, Plus 3 equals zero. So I am trying to see if this will have some factors. I am not too sure. So your product of the a and c is going to be 15. And the sum that you want is going to be 10. Is that possible? 8 and 2 will give you 16. That’s the closest you can get. Yeah. So this, 7 and 3 will become 21 and after that it will keep going somewhere else right. So these are non-factorable so which means you have to go for a, writing it using a quadratic formula. So we will have t equals minus b plus or minus, right. So it’s, minus b is 10 plus or minus root of b square minus 4ac, b square is 100 minus 4 into 15, ac is 15 right, 5 into 3, yeah, b square minus 4ac by 2 into a. a is your 5. Yeah? So what you will get with this now is, see now, this shows that not always will we get pretty looking answers right, so when it happens, that happens also we have to keep our heart in place. Not start doubting our solution. We have done everything correctly till here so even if the answer looks like it has a root, it must be correct. So you have 10 plus or minus something. This 10 and this 10 can cancel, so your time will be 1 plus or minus root of 100 minus 15 into 4 is 60 right. Root of 40 by 10. This is where you are.