 Myself, Ms. Shailaja Dehavarkonda, Assistant Professor in Civil Engineering Department, Vultured Institute of Technology, Solapur. In today's lecture, we are going to study the inclined projection of the projectile on leveled ground. At the end of this video, the viewers will be able to analyze motion of freely projected particle and to determine the motion of freely projected particle in inclination. Now consider the motion of projectile inclined means if you observe this figure here the particle A is projected inclined with the horizontal surface with an initial velocity u. Now the particle have motion in vertical as well as in horizontal because it will have in inclined projection. So whenever it is inclined it will have two directions that is horizontal as well as vertical. So here we will get two motions in vertical as well as in horizontal. So first of all we will calculate the vertical motion whenever you will consider the vertical motion of body the initial velocity is u sin alpha but as the gravitational acceleration is 9.81 meter per second square. It is acting downwards so here you will get minus 9.81 meter per second square. Now by considering the horizontal motion the horizontal component of the velocity is u cos alpha. So by these two equations we can identify or we can calculate the equation of trajectory. Trajectory means it is the path traced a curved path traced by the particle. Now let P represent the position of projectile after t second. If here you observe this figure P will represent the position of particle after the interval of time t. Now considering the vertical motion y is equal to u sin alpha that it is nothing but the motion of equation of linear motion that is ut minus one half gt square. So when you put the value of initial velocity in the equation you will get u sin alpha into t minus one half gt square because the gravitational acceleration is acting downward. And when you consider the horizontal motion of particle you will get the horizontal component as x is equal to u into t that is u cos alpha into t. By this u cos alpha into t you will get the value of t means it is t is equal to x by u cos alpha. By substituting the value of t in equation of trajectory you will get the equation as shown here. So after simplifying this equation you will get the final equation of trajectory that is y is equal to x tan alpha minus one half into x square g divided by u square into one plus tan square alpha it is the final equation of trajectory. Now to find out the maximum height of the particle we will consider first vertical motion. Whenever you all consider vertical motion it will be vertical component of the particle that is u sin alpha where when the particle reaches at maximum height if you observe when the particle reach at this point c because it is the maximum height of particle after which it is going downward or it will reach the ground. So when the particle is at its maximum height the final velocity will be zero. So here the initial velocity is u sin alpha and final velocity is zero and the acceleration is minus g because it is acting downward. So now by considering the equation of linear motion that is v square minus u square is equal to 2 as here s is denoted as the maximum height and a is the gravitational acceleration as we know when the particle reaches its maximum height the final velocity will be zero. So by putting these values in the equation you will get h is equal to which is nothing but the displacement s. So you will get here h is equal to u square sin square alpha by 2g. So by using this equation you can find out the maximum height of the particle. Now you have to find the time required to the particle to reach maximum height by using first equation of motion that is u is equal to u plus at when whereas v is the final velocity u is the initial velocity and a is the acceleration. When the projectile reaches maximum height you will get final velocity is zero initial velocity is u sin alpha minus gt. So by this equation you can find out the time required to the particle to reach the maximum height. So which is calculated by t is equal to u sin alpha by g. Now motion of projectile in vertical direction is given by the equation of trajectory that is y is equal to u sin alpha into t minus 1 of gt square it is nothing but the equation of trajectory. At the end of flight y is equal to 0 means vertical height is equal to 0 you will get into u sin alpha minus 1 of gt square when the t is 0 gives initial position of o of the particle. So by using this equation the time of flight is calculated as t is equal to u sin alpha by g. So hence the time of flight is calculated as t is equal to 2 u sin alpha by g. Now horizontal range which is nothing but the distance horizontal distance of the particle to reach the ground level. During the time of flight the projectile moves in horizontal with uniform velocity u cos alpha because the horizontal component of the inclined particle is u cos alpha. Hence the horizontal distance travelled by the projectile is given by r is equal to u into t whereas u is the initial velocity. So here r is equal to u cos alpha into 2 u sin alpha by g it is the value of time of flight which is we have calculated above. So finally you will get the horizontal range is r is equal to u square sin 2 alpha by g. So by using this equation you can calculate the horizontal range of the projectile. Now by using these all equations of trajectory time of flight maximum height and the horizontal range you have to calculate the positions or parameters of the projectile. Now the problem is given you pause the video and try to solve the problem. In this a body is projected at an angle such that its horizontal range is three times the maximum height find the angle of projection. So by using above equations you will try to solve this problem and answer this. Now this is the answer for the problem. These are the references are carried out for the further study. Thank you.