 Pratik stands on top of a building and throws a stone upwards with a speed of 25 meters per second. The stone falls to the ground in 10 seconds. Find the total distance travelled by the stone and we can take G as 10 meter per second square. Why don't you pause the video and first attempt this one on your own. Alright, hopefully we have given this a shot. Now let's try and draw what's happening in the question. So we have someone standing on the top of the building and throwing a stone upwards at a speed of 25 meters per second. Now this stone falls to the ground in 10 seconds. So it goes up and then it comes back again and falls to the ground and the time it takes to do that is 10 seconds. We need to figure out the total distance travelled by the stone. So if we label that over here, that would be this distance right here. We can call this D1 and this long distance right here. We can call this D2. So we need to figure out D1 plus D2. This is what we need to work out. Now we can try approaching it by breaking the motion into two parts. One part of the motion is when Pratik throws the stone and the stone reaches the top height right over here and then the second part begins after it starts falling until it reaches the ground. So we can treat this part separately. We can use, we can see equations of motion, try to figure out D1 and then we can treat this long part separately, try to figure out D2. So let's see what all do we know. Let's try to list down all the variables and see if we can work out D1 and D2. For the first part of the motion, what all do we know, we know that the initial velocity that is U, let's write that as U, that is 25 meters per second. And final velocity, at the top most point, we can say that the final velocity is zero because the stone momentarily comes to a state of rest. There is acceleration acting at the stone, therefore it comes down, it starts falling down, but at that point, at the top most point, the stone is at rest. So the final velocity for the first part of the motion, we can take the final velocity to be as zero. So V is just zero. And what else do we know? We know that there will be some acceleration, in fact, there will be acceleration due to gravity, which would be acting down. And if we try to fix a coordinate axis, if we try to fix a reference frame, let's take the reference frame of Pratik. So from Pratik's reference frame, we can say that there is a y-axis going on the top and there is positive x going to the right. This is positive y, this is positive x. So the acceleration vector is pointing down, that is in the negative y direction. So acceleration here, this is minus g, and that is minus 10 meters per second square. So we know three variables, that means we can use an equation of motion. We need to know at least three variables to be able to use one. So in this case, in the first part of the motion, we can use v squared equals to u squared plus 2a into d1 into d1. We know that v squared is zero and u squared, this is 625. So zero, this is equal to 625, 25 squared plus 2 into minus 10. So that is minus 20. So let's write this as 2 into minus 10 into d1. This is multiplied with d1. So d1, that is equal to 625, and when 625 goes to the left-hand side, it becomes minus 625. So that minus gets cancelled with this minus over here. And d1, this becomes equal to 625 divided by 20. And when you work this out, this comes out to be equal to 31.25 meters. So d1 is 31.25. Let's come to the second part of the motion and figure out d2. Again, let's try to list down all the variables that we know for this part of the motion. So for this one, we know that the initial velocity is zero. It's starting, we are considering the second motion from the top post point to right to the very bottom. So initial velocity, that is zero. Initial velocity is zero. Final velocity, we don't really know. So okay, let's try the question mark for that. We know acceleration, acceleration is still g, it's 10 meter per second square, right? And still, acceleration is again acting downwards. Even though it is in the direction of the motion, it is still acting downwards, which is in the negative y direction, and acceleration is a vector. If it's pointing in the negative direction, there is a minus sign attached to it. So it's minus 10 meters per second square. We are in the reference room of Pratik, right? So negative y is in downwards direction. And what else do we know? We don't really know anything else. But we can't use an equation of motion because we only know two variables. We know u and acceleration. So we can't really work out d2. But the question also tells us one more thing. It tells us that the stone falls to the ground in 10 seconds. That means that from the time the stone leaves Pratik's hand, till it reaches the ground, it takes a total of 10 seconds to complete this journey. So if we figure out the time that the stone took to cover the first part of the journey, then we can subtract that from 10 to figure out the time that it took for the second part of the journey. So time for the second part, this really is 10 minus the time for the first part. And we can actually work out the time that the stone took for the first part of the journey because we at least know three variables over there. So let's do that. Let's do that. And for this, we can use this equation v equals to u plus at because we need to work out time. So we are using this equation. Final loss t is still zero. This is 25 and acceleration is minus 10. So 25, when it goes to the left-hand side, this becomes minus 25 equals to minus 10 into t minus gets cancelled off. And time comes out to be equal to 2.5 seconds, 25 divided by 10. So t2, this is actually t1, the time that the stone took for the first part of the journey. Time for the second part, t2, this comes out to be equal to 10 minus 2.5. This is 7.5 seconds. So the stone took 7.5 seconds to cover this distance of d2. Now we can work out d2. We can use this equation because we know three variables. So we can use an equation of motion. So let's use that one, which has acceleration, time, distance and initial speed. So that one is s is equals to d2 s that is equal to ut plus half 80 square, ut plus half 80 square. Because we know three out of these four variables, u is zero. This is zero. So this whole factor, this becomes zero. Half 80 square, this becomes equal to, this is half into minus 10 into 7.5 whole square. And when you work this out, d2 comes out to be equal to 281.25 meters. Yes, there is a negative sign before that. But because we are figuring out the distance, we can use the mod sign. So really, we need to just figure out, we need to figure out d1, mod of d1 plus mod of d2. We can use the magnitude. We can use its magnitude. If we were figuring out the displacement, then the story would have been different, right? Then we would have worried about the negative sign. But because we are calculating distance, we only need to just take the positive value. We just need to take the magnitude of the value. Don't have to worry about positive or negative sign. So this is d1 that is equal to 31.25, 31.25 plus 281.25. So the final distance, this comes out to be equal to 312.5 meters. This is 312.5. You can try more questions from this exercise in the lesson. And if you are watching on YouTube, do check out the exercise link which is added in the description.