If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:11:12

Video transcript

- [Instructor] Let's solve a problem on two objects in free fall. Here it is, a stone is dropped from a height of 100 meters and at the same time another stone is thrown up with 50 meters per second from the bottom. Find when and where do the two stones meet. So, let's quickly look at what's given and try to make a drawing. A stone is dropped from a height of 100 meters, so there's one stone that's dropped from 100 meters. And at the same time another one is thrown up with 50 meters per second from the bottom. And we need to find when and where do the two stones meet. So, if one stone is falling down and the other one is thrown up, then they could meet somewhere, right? So maybe this stone falls down here, and in that same time maybe this goes all the way up to here and maybe it meets over there. They're gonna hit each other at some point. We need to figure out when, that means we need to figure out after how long after we have thrown and dropped them. So a time is what we need to calculate, and we need to also figure out where, where exactly, what point is this. So, maybe we need to calculate the height. Either we can calculate this distance, or from the top, anything is fine, we'll calculate the height, let's say. So, these are the two things we need to calculate. So, how do we do this? Well, we know that whenever things are falling, they have a constant acceleration, and as a result they should obey the three equations of motion. And so maybe we can use these equations for both these stones and then see if we can calculate h and t from them. All right, so let's do this, let's call this our stone number one and this is our stone number two. Let's make a division because we will have two equations for both the stones. So, let's target stone number one. Here's stone one. Let's look at one data, what is the one to us? What do we know about this stone? Well, we know it's being dropped, and that means, whenever you're dropping something, it means that it has an initial velocity zero. So, we know u is zero. We know its acceleration, its acceleration is plus 10 meters per second squared and it is plus because this stone is, as it falls down, its velocity increases, and as a result, acceleration is positive. If velocity were to decrease, acceleration would be negative. Okay, t is something that we don't know, we just call it as t, we need to calculate that. What else? What about its displacement? Well, we are taking this point at which the two stones are meeting as h, that height is h. And right now the stone has a height of 100 meters, so, in time, t, when they meet each other, how much will be the displacement of that stone from here to here. What will this length be? Well, that is going to be 100 minus h, right? And so, its displacement, we can write, is 100 meters minus h. What else? What about final velocity? We have no clue about its final velocity, we don't care about it, so let's forget about the final velocity. So, we have these four things, now the question is, which equation can we use to connect these four things. So, why not give it a shot yourself? Can you try and pause the video and see which equation, which of these three equations would you use to connect these four quantities. Go ahead, give it a shot. All right, let's see. We won't use the first equation because it has a V in it and we don't have V. We wouldn't use the third equation for the same reason, there is a V in it. That means we have only one equation, so let me just cut those equations out. So, we have only one equation, S equals ut plus 1/2at squared So, let me go ahead, directly substitute in this equation and see what we get. So, if we substitute, S is 100 meters minus h, that equals ut, u is zero, so this term goes to zero, plus 1/2 a, a is 10 meters per second squared times t squared. And if we simplify that, we'll get 100 meters minus h, that will equal, 1/2 times 10 is five, so we'll get five meters per second squared-- Oops no square here. Times t squared. And we cannot solve this equation further, simply because there are two things that we don't know, we don't know h, and we don't know t, so we can't solve it. So, this is where we have to leave this equation. So, what else do we do? How do we calculate h and t? Well, we have the second stone. Maybe we'll build a second equation, and maybe that equation will help us. So, again, a great idea to pause and see if you can write the initial data and build the equation for the second stone yourself. Go ahead, give it a try. All right, so, for stone two, let's write the initial data. What is its initial velocity? Now it's not zero, it is thrown up with 50 meters per second. What about its acceleration? Well, since it's thrown up it slows down as it goes up, right? And because it's slowing down its velocity's decreasing, that means its acceleration is negative 10 meters per second squared. What else? What about its displacement Well since the stone goes from here to here its displacement is just going to be h. And time is this t, and again we don't care about its final velocity, we don't want that at all. So, which equation will be used to connect them? Well, the same equation, for the same reason, right? We don't want final velocities, and so if we substitute we get S which is h in our example, or in our case. h equals ut, that's going to be 50 meters per second t plus 1/2 at squared. And if you simplify this again, minus 10 divided by two, you get a minus five. So, let me just quickly write that down, so we'll end up with h equals 50t minus five t squared. And again we get an equation with two unknowns, and so we can't solve it. But, but, now we have two equations with two unknowns, and you may have already learned in maths that when you have two equations with two unknowns we can solve it. How do we solve things like this? Well, we can substitute for one variable into another. For example, maybe from this equation we can calculate what t is and substitute in this equation. Or, maybe we can calculate what h is from one equation and substitute in another. So, you know what? Since I already know what h is over here, let me substitute this value in this equation and see what we end up with, all right? So, let me write this equation down. So, if I write this equation down I'll get 100 meters minus h, which I'll substitute from this equation, so let me just go ahead and copy that part, and paste it over here. So that's my h, so let me put that in bracket. Okay, that should equal, five, five meters per second squared times t squared. And now, look, we only have one variable, only t. And so if we solve this, we can now calculate what t is. And once we calculate what t is, well, we can use these equations and calculate what h is. So again, it would be a great idea to pause and see if you can try this part yourself. Now, it's purely mathematics, the physics has ended. Okay, anyways, let's do this. So, let me scroll along because I don't need any of those things. Okay, so, if you further simplify, let's open up this bracket, we'll get a negative here and this will be positive because a negative times a negative will be a positive. So again, if I quickly write this, I get the same thing, I get a positive here, and everything else remains the same. And now look! We have a five t squared here and we have a five t squared over here, so that just cancels out. We can subtract five t squared on both sides, and we are lucky because the t squareds cancel out, otherwise it would've been a quadratic equation, there would've been more steps. But now because the t squareds are gone, that means we are left our 100 meters minus 50 meters t equals zero, equals zero. And we can now solve it in a couple of steps. So if I add 50t on both sides then I'll end up with 100 meters equals 50 meters per second times t. And, if we divide by 50 meters per second on both sides, then we get what t is, so the zero cancels, five goes two times, the meter cancels, and we'll end up with a two second on this side. So, we found what t is, t equals two seconds. That means after two seconds of dropping and throwing up that stone they will meet each other. Now, before we go back up there, let's calculate what h is. How do we do this? Well, now that we know what t is, we can substitute in any of these two equations and I'm pretty sure you can do that. You just substitute the 2 in any of these equations and calculate h, and if you do that, which I'll leave it to you, to save time, if you do that you will end up with h equal to 80 meters. And so let's zoom out and put everything in one frame. Yeah. There it is. And so whenever we have two objects in free fall and we're asked to calculate when or where they meet, what do we do? Well, we build equations for both of them by using the equations of the motion. And then we solve the two equations to calculate the height and the time.