 Hello everyone, myself, Mr. Elaja Dhavarkonda, assistant professor in civil engineering department waltz instead of technology, in today's lecture, we are going to study kinematics of particle. Learning outcomes at the end of this session, the students will able to calculate the velocity and direction of the projectile for different levels. The starting point of the projectile and the strike point of the projectile is at different height. When the particle is projected on level ground surface, it will strike on the same level. The point of projection and point of strike is different in this case. If you consider any particle and if you are throwing it upward, normally we will observe it is going vertical, but it will have the two directions that is vertical as well as horizontal direction. So we will consider it as the inclined projection. So in this lecture, we are going to study the point of projection and the point of strike at the different level. Now let us consider the point of projection is at height of y0 above the point of strike as shown in figure number 1. So here you can observe this figure here. The particle is thrown or the projection of the particle is inclined with the initial velocity u. After reaching a certain point, the velocity of particle will be 0 and again it will coming downwards to the earth due to the gravitational force. So here you can observe the starting point of the particle or the projectile and the strike point of the projectile are at different level. Normally this case is observed in the sloping grounds and all. So the difference between the point A and B is shown or is notified by the y0. So it is the height between the starting point and the strike point of the particle. This point C is the maximum height which is reached by the particle. Already we have seen the projection of projectile on the level ground surface. Level ground surface means the starting point and the strike point is at the same level. In that case the horizontal range is calculated by u into t means it is initial velocity into time required to reach from point A to point B and the maximum height is given by one half g t square. G is the gravitational force and t is the time required to reach the maximum height. So in this case we will find the velocity where the point of application and the point of strike will be different. So by considering the vertical direction of the projectile for a same ground level, the y is equal to u sin alpha into t minus one half g t square. So it is the horizontal component of the projectile. On the equation of vertical direction of the inclined projectile you will get y is equal to u sin alpha into t minus one half g t square. This u sin alpha t is the vertical component of the inclined projectile. Here the strike point is below the point of application. So we will consider here y is equal to minus y 0 because it is below the starting point or you can say it is below the original level. So the time required to reach at station B is given by minus y 0 is equal to u sin alpha into t minus one half g t square which is and the horizontal range is given by r is equal to u cos alpha t into and the horizontal range is given by u cos alpha into t. This u cos alpha is the horizontal component of the inclined projectile. So here they have mentioned the height and you have to calculate the time required to reach the next point. Here they have mentioned the height and you have to find out the time required to reach the end station. Now we will see the problem how to calculate the time of light range and maximum height. In this problem a bullet is fired from a height of 120 meter at a velocity of 360 kilometer per hour at an angle of 30 degree upwards means it is inclined projection throwing upwards with the inclination of 30 degree. So neglecting the air resistance find the total time of flight, horizontal range of bullet, maximum height reached by the bullet and final velocity of the bullet just before touching the ground. Now we will see further how to calculate these things. The initial velocity is given as 360 kilometer per hour. So if you convert it it will be 100 meter per second. The angle of projection is also given in problem that is 30 degree. So this initial velocity is in kilometer per hour. So first you have to convert it into meter per second. So it is initial velocity is 100 meter per second. Then total time of flight we have the equation that is by considering the vertical motion y is equal to vertical component of the projectile that is u sin alpha into t 1 half g t square. So here the height is given as 120 meter means the strike point is given as 120 meter. So it is below the point of application so we have considered it as a negative. Initial velocity u is 100 and gravitational force is given as 9.81. So putting these all values in this equation y is equal to u sin alpha t minus 1 of g t square you will get the time required to reach from point A to point B. So to reach the particle from point A to B is 12.26. Then you have to calculate the maximum height reached by the bullet. The maximum height is calculated by u sin square alpha square. Next you have to find out the maximum height reached by the bullet. Maximum height means this bullet is reached at this point C means it is the maximum height where it will go with its initial velocity and at this stage at this point C its velocity will be 0 and again it will start flowing down. So this maximum height is calculated by h is equal to u square sin square alpha by 2 g. So these all equations are derived from the inclined projection on the leveled ground. So initial velocity is 100 by putting the yellow of initial velocity and gravitational force you will get the maximum height that is 127.42 meters. So this 127.42 meters is from the point A but your bullet is strike below the 120 meter from the point A. So the maximum height should be calculated from the point B. So here the total height should be 127.42 plus 120 which is 247.42 meter from the ground where the bullet is reached or where the bullet is strike. Now you have to calculate the horizontal range. Horizontal range means it is the distance between horizontal distance between the two points means start points and end point. Due to inclined projection you have to consider the inclined component of the projector. So it r is equal to for the vertical component sorry for the vertical projector you will consider u into t that is initial velocity into time. Here it is inclined so we are considering horizontal component of this inclined projection Now you have to calculate the horizontal range. Range is nothing but the distance between two points that is start point and end point. Here this is the inclined projection of the bullet so you have to consider the horizontal component of the bullet. So it is u cos alpha t alpha into t. So t already we have calculated that is 12.20 seconds and initial velocity is also known as that is 100 meter per second. So horizontal distance between point A and B is 1056.55 meter and the total height is 247.42 meter from the point B. Now we have to find out the final velocity just before striking the ground. Here it has the bullet is projected inclined so it has vertical component as well as initial component. So first we have to calculate its vertical component and initial component. So here the vertical component of the velocity is u sin alpha t sorry u sin alpha minus g into t. So the vertical component of the velocity is 69.628 meter it is negative means it is representing as it is acting downward because we have to calculate the velocity just before to strike the ground. So it is gravitational force is acting downwards so we have considered here it as a negative. So this velocity is acting downward. Then horizontal component that is u cos alpha so you will get the horizontal component of velocity is 86.603 meter per second. Then the final velocity will be nothing but the resultant of these two forces. So the resultant is calculated as some fx under root of fx square plus fy square. So the same formula is applied here and have calculated the final velocity. See this is the inclined projection so we have calculated the horizontal and vertical component. From component of velocity and from that we have calculated the original velocity of the bullet. So this bullet is strike to the ground with an velocity of 1111.16 meter per second at an inclination of 38.82 degrees. So likewise we have to apply the equations to the given problem and we have to find out the unknowns. That is time of flight, maximum height, horizontal range or either velocity. Now you pause this video and try to answer this question. These are the references considered for the study. Thank you.