 Okay, so we have taken type 1 project time, last class I am talking about, we have taken type 1 project time. Okay, what else we have here? The velocity is going to be in front of us. Okay, to find out. Understood. Now if I want as my y coordinate and this as my x coordinate, but initially 0, right? So finally what is x equals to? After time t, the value of u cos theta into t. So this is x. So this is going to be u sin theta. Okay. So now I have u cos theta. Substitute that here with y equals to x. The degree of 1 and degree of 2. So this kind of equation represents parabola. You will learn about past coordinate geometry. But it is good to know that. So you can see that when you put this equal to 0, you get x max which is equal to your range. Right? You can see that yourself. Right? So this is the result. Minus half s square by let's say t. Okay. And then what? Use g divided by, g divided by 1 by 2. You can substitute this value and value of g. You will get the value of shens in which equation of path is given to you. Not the initial velocity and initial angle. Then you have to correlate with the equation of project time. What? No. How? For g is already there. Sorry. g need not be there. So 1 divided by r is equal to 1. Understand? Understood? So do not remember the function for every one. Equation s equal to up plus half it is along x axis and along y axis. Just write it. Don't solve anything. Along x and y axis write down this equation. Initial velocity x axis is what? Step number one. Decide where it is 0. Why? Because y direction is per line. So initial velocity along y axis. Acceleration along x axis is what? And along minus g. Now if I take point 1 and point 2. What is the displacement along x axis? R. R. R? Displacement along x axis is R? Alright. And displacement along y axis is what? Minus x. Minus x. It is down. You are taking all the vectors. Substitute these values. Minus h is equal to minus half gt square. Yes or no? Between point 1 and point 2. Alright. So luckily this y axis equation will give this under root 2x by g. Now this time of flight something strange about it. What is that? It will fall like this. Throw it with higher velocity. Small deviation because of air drag. But air is not. So this is my. So the range I will find out. Velocity is the kind of the root 2x by g. Because t I have got from the y axis equation. So you can get the value of 2x. Understood. This is time 2 vector which is seen in the final answer. Alright. So even if you don't remember and that is point 2. By the way what was the biggest question? Naturally because it is vertical axis on its own. But here it is not and then along y axis. Do it quickly. Done. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. Sir. be in that mode which I am telling you what people think minus h if it's simple change in y coordinate is minus half dv square so I am now telling you one thing how will you find the maximum height to maximum height it's your height do it now it's attempted find out the maximum height this is how much it is see you so v by is the u square sin square theta minus 2g into into what it's y coordinate so y coordinate will be u square sin square theta divided by 2g now this is my u square sin square theta divided by 2g after that y coordinate will be 0 what? I will use this equation between so v by is u sin theta and equation is minus g then s by is equal to 3 what? why didn't you take s as y? this is the basis s y this is s y so here there will be minus u sin theta into t it's going down