 Hello friends, Myself Sandeep Javeri, Asgenu Professor, Department of Civil Engineering from Valshan Institute of Technology Shwalaapur. In today's session, we are going to discuss problem on linear motion for the vertical type of motion. The learning outcome at the end of this session, students will be able to solve problem on linear motion for vertical type of motion. Now let us consider linear motion. The motion of a body in a straight line is known as linear motion. For example, a car is moving on the road, so it is moving in a horizontal plane. The second example, a body is projected vertically upward in the air with a certain velocity that means the motion is happened in vertical plane. The third example is a body falling vertically downwards, so these are the examples of linear motion. Now, let us consider the equations of vertical motion with uniform acceleration. Type 1, a particle is moving vertically upwards. The expression for the final velocity v is equal to u minus gt, where g is an acceleration due to gravity, u is an initial velocity and t is a time taken during the motion. Here minus is sign because the direction of g is towards the center of the earth, so here in vertical upward journey, here the minus sign is taken. Now, the second expression to find the height h, h is equal to ut minus half gt square. Similarly, third expression v square is equal to u square minus 2 times g into h, where h is a vertical displacement or height and u is initial velocity, v is a final velocity and g is an acceleration due to gravity. Now, let us consider type 2, when particle is moving vertically downwards, here expression for final velocity v is equal to u plus gt, here plus sign is due to the acceleration due to gravity, g is in a downward journey, it is acting towards downward direction, so it is plus, where second expression h is equal to ut plus half gt square, third expression v square is equal to u square plus 2 times g into h, where h is a vertical displacement or height, u is initial velocity, v is a final velocity and g is an acceleration due to gravity. So, the difference between type 1 and type 2 is, we are taking in type 1 g as minus, in type 2 g as positive, because the direction of acceleration due to gravity is towards the center of the earth. Now let us consider the problem on vertical motion with uniform acceleration, a body falling freely under the action of gravity passes two points, which are 20 meter apart vertically in 0.4 seconds, from what height above the higher point, did the body start to fall, so take g is equal to 9.8 meter per second square. Now let us solve this problem, this is a type of vertical motion problem, a body let us say start from a and it moves vertically downward and it reach to point b, now again it reach to point c, the distance from b to c is given as 20 meter and during journey from b to c the time taken is 0.4 second, so the question is from what height above the higher point did the body start to fall, so total height we have to find out. Let us consider the x is a distance from a to b, now the time taken is say t second from a to b, so we know that when the body is moving vertically downward, so the initial velocity at a is equal to 0 meter per second, as the motion is vertically downward, so accession due to gravity is positive, so g is taken as positive. Let us consider the solution for this, consider motion a to b, so motion a to b initial velocity is 0, so time taken is t second from a to b and distance from a to b is x meter, so it is x meter, so using the expression h equals to ut plus half gt square, so the distance from a to b is x meter x equals to, now initial velocity is 0, so it is 0 into t that is the time taken from a to b plus half g is 9.8 into t square, so we are getting x equals to 4.905 t square, this is the expression number 1. Consider motion a to c again, so consider motion a to c, so we have the total distance of a to c is x plus 20 meter and time is t plus 0.4 second, so this is the point b that is intermediate point b, this is the last point c where the body reach, now initial velocity is 0, now using the expression h equals to ut plus half gt square, h equals to x plus 20 meter which is equals to u that is initial velocity at a is 0, so it is 0 into t is a time total time is t plus 0.4 second plus half into 9.8 into time is t plus 0.4 whole the bracket square, so we are obtain the expression x plus 20 is equal to 4.905 into t plus 0.4 entire the bracket square, this is equation number 2. So, equation 1 is x equals to 4.905 t square, this is equation number 1, now solving equation 1 and 2, so we are getting t is equal to 4.92 second and by putting this value in this expression 1, we are getting x equals to 4.905 into 4.90 second bracket square, 4.92 square it is equals to 117.64 9 meter, so finally we are getting x equals to 117.649, and from a to b distance we are obtain as 117.649 meter. So, what is start falling from height of 117.649 meter above b and the total distance will be from a to c will be 117.649 plus 20 that is this total distance is 137.649 meter, you are suppose to pause the video and answer this question. So, this is the answer, now to solve this problem, we have to use the expression v square minus u square is equal to 2 g h. So, we are getting v is equal to under root 2 g h from that expression we can obtain what is the final velocity, these are the references that we are using for conducting this session. Thank you.