Projectile Motion & SUVAT - A-level & GCSE Physics

So hopefully you've watched my video on one-dimensional suvat using Newton's equations of motion. Now let's move on to two dimensions, that is, projectile motion. Okay, so let's say that Bob wants to have a go at some tombstoning, as they call it, so running off the edge of a cliff, and he wants to get as far as he can into the water. So here's Bob, right here. He's about to run off there. Bob's gonna take this trajectory into the water. Now I can tell you Bob is pretty fast, so he's running with a speed of eight meters per second. I want to know how far he actually gets. I want to know his distance there. I'm gonna call that D. Now the first thing we need to do is separate Bob's motion into vertical and horizontal components. Whenever something's thrown up into the air at whatever angle, we know that whatever's going on vertically with its speed and acceleration is gonna be completely independent of whatever's going on horizontally. Let's say accidentally he trips, and as opposed to purposefully tripping, he trips here and then he just, whoop, falls right down the cliff. So he lands right here. That's actually gonna take him the same amount of time, therefore showing that it doesn't matter how fast he's going, he's always gonna take the same amount of time to get down to the water. Okay, so let's write down what we know is going on suvat-wise vertically. This is how you lay out your question. So you write down your suvat as per usual. I can tell you that this cliff is 50 meters high, very, very high. He's a bit of a daredevil, is Bob. So therefore we can write down his displacement vertically is 50 meters. Do we know anything else about Bob's movement vertically? Well, we should do because like we saw in the last video, whenever something just is going horizontally or just starts dropping, we know that its initial speed, initial velocity rather, is zero meters per second. This is the bit where people get confused. They get confused between initial velocity vertically and the speed at which something is traveling horizontally when it starts falling. Whenever something is going horizontally off an edge, you can always say that its initial vertical velocity is going to be zero. We don't know V, so we're going to forget about that. Acceleration, well, he's falling under gravity, so we know A is going to be 9.8 meters per second squared. We're trying to find out time. So we need to find out what time is here. So we're going to use the equation S equals UT plus half AT squared. Now, just as an aside, this equation, you're only going to get given this equation if U is zero, if you're trying to find out T. If U wasn't zero, then you're going to end up with T squared times something plus T times something, and then you're going to have to use the quadratic formula to actually figure it out, but you don't have to do that at GCSE or A-level physics. So you're going to get this. You're going to have to use this to find out T only if U is zero. So we know U is zero, so we can get rid of UT there, and we need to rearrange it to find T. So let's do this one step at a time. 2S equals AT squared. So I can say 2S divided by A is T squared. Finally, that means square root in the whole lot gets us T is the square root of 2S divided by A. Pop that into my calculator. So I'm going to have the square root of two times 50 divided by 9.8. That gives me a time of 3.19 seconds. Great, so now that I know how long it takes Bob to actually get down to the water, I can actually use this time, and this is usually the way it goes. It's the time that links the vertical and horizontal components together. Because, of course, distance, speeds, and acceleration can be different, horizontal and vertical, but time is always going to be the same between the two. So I'm going to carry this over to my horizontal calculations now. Now, of course, we know that if something's traveling horizontally, it's not going to keep on going forever because, of course, you have air resistance. But we forget about air resistance because we just pretend it isn't there. No air resistance at all, no drag forces. Now, if there's no air resistance, then that means that there's no forces acting horizontally on Bob at all. The only force acting on him is his weight, pulling him downwards. Therefore, we can say that speed is constant. If speed is constant, then that means acceleration is going to be zero. So actually, that means that there's no need for SUVAT. So that's the good thing about projectile motion at GCSE and A-level. You never have to use SUVAT in both vertical and horizontal components. You only have to do it vertically. So if you know speed is constant, then we can just say speed equals distance over time. That means, therefore, the distance is speed times time. And of course, when we talk about speed, we're talking about its horizontal speed. That horizontal speed, it started off at eight meters per second. That's always going to be the same. So we're going to say eight times 3.19. That gives us a distance of 25.6 meters. Therefore, we can say that D is 25.6 meters. That's how we do projectile motion. We use SUVAT vertically to find out the time. Usually, this is the way it goes. And then we use that time to put that into our speed equals distance over time equation and find out a distance. Alternatively, you could get given the distance and the speed here to figure out the time and then pop that back into your SUVAT equation. So just be aware, it's the time. It's the only thing that's going to be constant between horizontal and vertical components of the motion. What if you were asked, what is Bob's final speed, final velocity, as he enters the water? Now, when we're talking about his final velocity, we're not talking about after he's hit the water, but what speed he actually hits the water with. So of course, we're talking about the speed before he's slowed down by the water itself. Now, here's Bob as he's about to go into the water. Now, we know horizontally that his speed is eight meters per second. If I wanted to find out his vertical velocity as he enters the water, that's obviously gonna be our V. Now, we could use V equals U plus AT, but generally, I would suggest that because we've actually calculated what T is, then I wouldn't use that. Let's use the information that we've actually been given in the question so we can avoid as many possible problems. So we're gonna use V squared equals U squared plus 2AS. Of course, we know that U is zero, so that disappears. So V is gonna be the square root of 2AS. That's gonna be the square root of two times 9.8 times 50. So that gives us a final downwards velocity of 31.3 meters per second. So I can put that on there now. Instead of V, I can say that's 31.3 meters per second. We know his final velocity is obviously going to be in this direction. How do we find that out? Well, as per usual, like you've seen before, what we do is make ourselves a triangle if you really want to so we just pop this around here and we have ourselves a triangle. All we have to do then is use Pythagoras. Let's call this speed, I don't know, 18 that we've already used V. So A is just gonna be the square root of A squared plus 31.3 squared. That gives us a final speed of 32.3 meters per second. What about if you wanted to find out what angle he hits the water at? Well, technically, the angle to the horizontal is gonna be that there. Let's call that theta for now. And of course, that's going to be the same there. So we have our triangle and obviously we have our adjacent and we have our opposite so therefore theta is gonna be equals to the inverse tan of our opposite divided by our adjacent. That's gonna give us a fairly big angle of 75.7 degrees. This is just for A level now. What if we had something actually being fired up at an angle? Let's say we've got a cannon here and it's gonna be firing a shell upwards at an angle of 30 degrees. Let's say it fires it at a speed of 80 meters per second. Of course, as per usual, we have to split this into horizontal and vertical components in order for us to do anything useful with it. Obviously, we're not going to have components that are bigger than 80 meters per second. They're always going to be smaller. So we're gonna have to times by cos of 30 or sin of 30. But which way round is it? Well, of course, like from our rule in our previous video, if you turn through the angle, then we use cos. If we're turning away from our angle, we're turning away from our sin, turning away from our sin. So this horizontal velocity as it leaves the cannon is going to be 80 cos 30 and then the vertical one is going to be 80 sin 30. So we can say u horizontal is that, u vertical is that. Now, of course, if there's no air resistance, this is going to stay constant and we don't have to use SUVAT horizontally at all. So therefore, that's gonna stay the same. But then vertically, obviously, we've got to do something with that. This speed, it turns out, is 40 meters per second. Now, of course, if we're saying that upwards is positive, downwards is negative, when it comes back down the other side, so obviously it's taking this trajectory here, by the time it reaches the other side, we know that its final vertical velocity is going to be minus 40 meters per second, just like we saw with me throwing the ball up in the air in the previous video. You have your u, you have your v, you know what a is, therefore you can find out what time and the displacement are going to be. Obviously, the displacement vertically is going to be zero so that we don't really use that at all. What we usually do is find out the time it takes for it to go up and come back down, and then we use that as per usual to find out how far it's gone. Of course, a in this case is going to be minus 9.8 meters per second squared because it's accelerating downwards even though it's going upwards to begin with. So that's the difficult bit. As soon as you resolve your resultant speed, velocity into horizontal and vertical components, you treat it just the same as the previous question that we saw with Bob jumping off the cliff. So there you go. I hope you found this video helpful. If you did, then please make sure you leave a like and make sure you leave a comment as well down below suggesting what I can cover next, and I'll see you next time.