Second case and that is horizontal projection. From a building of H height, we have projected a particle horizontally with a velocity of u. Path followed by the particle is parabolic and the motion is projectile. We will take this direction as x and this direction as y, and g is acting downward. For this projectile we had to develop many things, time of flight, horizontal range, velocity at time t and equation of trajectory, velocity of impact with the ground. This is question of projectile motion, and is it clear. For projection motion what is the first step. We have to make a table. xy ux uy ax ay sx sy. Positive and negative we already are clear about that, how do we take that. Okay? I will ask and you answer.
ux
[students answering]
initial horizontal velocity. What will be the value?[students answering]
u or u cos theta?[u]
I think it should be u. How many of you have written u cos theta. I will solve your problem, don’t worry, you do make mistakes in the beginning. Now see, if this particle, now listen very carefully no talking, if this is thrown like that, this angle is theta and what will be this velocity component? U cos theta. Done? Now compare this with this. What can you make out? What is the theta of this case?[zero degree]
Isn’t it zero degree, tell me? Isn’t the theta zero degree?[yes]
So if we use u cos theta then we have to see that for this case theta should be zero degree and what is the value of cos zero degree?[1]
1, so then effectively what will be the value of this u cos theta, equal to u, will it not be? So what will be answer?[students answering]
And what is uy?[zero]
If you had thought of it as u sine theta angle then what will the theta be? Zero degree, that means sine zero zero, so initial velocity is zero.
ax horizontal acceleration.[zero] Vertical acceleration.[g] Now should we take g or minus g?[g] We had said that we would decide the sign of acceleration and displacement, by seeing the property of motion. Is it coming down only?[yes] So displacement and acceleration both are they in the direction of motion?[yes] So, it should be taken as g. Take plus g why do you want to get into the hassle of taking minus g. minus g is not wrong, it is not wrong. Okay, now it will be made more clear. Horizontal displacement is equal to range or no?[yes] We don’t know R. What is vertical displacement?[H] Now those who are writing minus g what should they take displacement as?[Minus H] Minus H, otherwise there will be a problem, because here you wrote minus g without thinking and then you have written H here without thinking, then you will have a problem. Either take both minus or take both plus. I will take both plus because I have learnt that I should stay away from minus. See vertical motion is downwards, that is why acceleration and displacement both are in the direction of motion and it should be positive. Okay.
See, the horizontal motion and identify. Again in horizontal direction ball is moving with constant velocity. Yes or no? Any doubt? Done?
What is vertical motion? Do one work, delete horizontal motion, horizontal has disappeared. Now we just have to see what is this motion?[free fall] What is the initial velocity in free fall also?[zero] The particle was dropped with initial velocity. Will that height H drop down or no?[yes] I remember that we had written two TNT for this. If the particle is dropped with initial velocity so how much time will it take to cover the distance H?[students answering] Root of 2 Hyg, do you remember?[Yes, sir] That’s why I had told you that both the TNT was very important. See the application has now come. I had also told you one more thing. After dropping H distance the velocity…[root 2gH] Root 2gH, now you should understand yourself why I had said these things. I had told you that time that it is infinitely important for every application. Now just see what happens, see what happens in future. In every topic the application of kinematics will be there. Yes or no? So the direction of y motion we actually know this. Done? Should we do the calculations? Any doubts?
Okay, let’s go for the first part.
Can you imagine there was a question in IIT this year. In J Advanced there was a question which asked to use this formula. The range formula which we will now work on, that had to be used to give the answer. In this year IIT J Advanced. It was match the column. It is just a simple evidence to justify my statement that these all things are extremely important. Put a lot of effort in kinematics. Do the best level of efforts in physics. The one who puts in a lot of effort in kinematics will not have to put in so much of an effort in future, I am telling you, you will see.
Time of flight….
ax horizontal acceleration.[zero] Vertical acceleration.[g] Now should we take g or minus g?[g] We had said that we would decide the sign of acceleration and displacement, by seeing the property of motion. Is it coming down only?[yes] So displacement and acceleration both are they in the direction of motion?[yes] So, it should be taken as g. Take plus g why do you want to get into the hassle of taking minus g. minus g is not wrong, it is not wrong. Okay, now it will be made more clear. Horizontal displacement is equal to range or no?[yes] We don’t know R. What is vertical displacement?[H] Now those who are writing minus g what should they take displacement as?[Minus H] Minus H, otherwise there will be a problem, because here you wrote minus g without thinking and then you have written H here without thinking, then you will have a problem. Either take both minus or take both plus. I will take both plus because I have learnt that I should stay away from minus. See vertical motion is downwards, that is why acceleration and displacement both are in the direction of motion and it should be positive. Okay.
See, the horizontal motion and identify. Again in horizontal direction ball is moving with constant velocity. Yes or no? Any doubt? Done?
What is vertical motion? Do one work, delete horizontal motion, horizontal has disappeared. Now we just have to see what is this motion?[free fall] What is the initial velocity in free fall also?[zero] The particle was dropped with initial velocity. Will that height H drop down or no?[yes] I remember that we had written two TNT for this. If the particle is dropped with initial velocity so how much time will it take to cover the distance H?[students answering] Root of 2 Hyg, do you remember?[Yes, sir] That’s why I had told you that both the TNT was very important. See the application has now come. I had also told you one more thing. After dropping H distance the velocity…[root 2gH] Root 2gH, now you should understand yourself why I had said these things. I had told you that time that it is infinitely important for every application. Now just see what happens, see what happens in future. In every topic the application of kinematics will be there. Yes or no? So the direction of y motion we actually know this. Done? Should we do the calculations? Any doubts?
Okay, let’s go for the first part.
Can you imagine there was a question in IIT this year. In J Advanced there was a question which asked to use this formula. The range formula which we will now work on, that had to be used to give the answer. In this year IIT J Advanced. It was match the column. It is just a simple evidence to justify my statement that these all things are extremely important. Put a lot of effort in kinematics. Do the best level of efforts in physics. The one who puts in a lot of effort in kinematics will not have to put in so much of an effort in future, I am telling you, you will see.
Time of flight….
