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Anderson Video - Projectile Motion Example: Impact Speed

Professor Anderson
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>> Hello, class. Professor Anderson, here. Probably getting tired of me saying that over and over again, but I enjoy it, so. Let's talk about an example problem. Let's do a projectile motion problem, and let's complicate it a little bit by putting you up on top of a building. Okay, so here we are on top of a building, and we're going to launch an object, and we're going to launch it up at an angle, okay? So, above the horizontal, we will call that angle theta. And now, as soon as we launch that object, it is a projectile, and so it's going to follow a parabolic curve. Okay. Something like that. And, let's say that we are starting at height h, and we're ending, of course, at ground level. And, let's give you a couple parameters here. We're going to say that the following is given. Vi equals 20 meters per second. Theta equals 30 degrees, and h is 45 meters. Okay? And, let's ask the following question. Question A is how long is this thing in the air? Question B is what is its final speed? All right. How do we attack this? Well, I wonder if we could do it, maybe, slightly backwards by attacking B first. Let's think about a different way to get the answer for B. We had four kinematic equations, and one of those kinematic equations looked like this. Vf squared equals Vi squared plus 2axf minus xi. Now, is that one going to help us? This is really Vxf squared and Vxi squared. That might help us if we knew a little bit about the acceleration. But, if this thing is a projectile, then we know that the acceleration is, in fact, equal to 0. And so, all this equation tells us is that Vxf squared is equal to Vxi squared or Vx final is equal to Vx initial. Right? It doesn't change its speed in the x direction. The x component of the velocity doesn't change. All right. But, let's rewrite this equation for y. Vy final squared equals Vy initial squared plus 2a sub y (yf minus yi). Hmm. What is ay for this projectile after it launches? Yeah? >> A negative 9.8 meters per second. >> Negative 9.8, right? As soon as we let go of the thing, it's just gravity. All right. So, we know that, and so we can calculate Vy final. Vy final squared equal Vy initial squared minus 2g times what? Well, what is y final equal to? Yeah? >> 0? >> 0. It's on the ground. y final is 0. y initial, we told you, was h. And so, it looks like we have everything we need here. It's 0 minus h, and so this simplifies a little bit. We have Vyi squared plus 2gh. Hmm. So, look. I have an equation here for Vxf squared. I have an equation there for Vyf squared. What we're interested in is V final. Okay? How do we do that? Okay, V final is somewhat in that direction. If we go back to our picture, right, it's coming down and to the right. Vx is going to be this component of that triangle. Vy final is that component of the triangle, right? We have some x component of our final velocity. We have some y component of our final velocity. And, now, look. This is another right triangle, and so we can say Vf squared equals Vxf squared plus Vyf squared. And, now we know all that stuff. Vxf squared is just Vxi squared. Vyf squared is Vyi squared plus 2gh And now, here's the really cool thing. Vxi squared was this. Vyi was that. And so, that thing is also a right triangle. Okay. This is when we launch the thing. And so, guess what? Vxi squared plus Vyi squared is just Vi squared. And, now, we just have to add 2gh, and guess what? We know all of those numbers, right? And so, now we can plug it in and see what we get. Vf is going to be the square root of Vi squared plus 2gh, and let's try the numbers and see what we get. So, Vi we said was 30. We've got a 2. 9.8 for g, and we said 45 meters for h. So, let's see if we can approximate that while you guys punch it into your calculator. Okay, so we've got square root of 30 squared is 900. 2 times 9.8. That's basically 20, times 45 is roughly another 900. And so, that is 1800 which is 18 times 10 to the 2, and square root of 18 is a little bit more than 4. 4 point something. I don't know. 4.3. I'll say 4.3, and that becomes 10 to the 1. So, I'm going to say this is 43 meters per second. Anybody have an exact number? Yeah, what'd you get? >> 42.21. >> 42 point what? >> 21. >> 21, okay. So, we were pretty good in our guess there. 42.21 meters per second for the final speed of this thing right before it hits the ground. Now, let's think about this for a second, and let's go back to this equation, right? We have Vf equals Vi plus 2gh, but remember, in our picture, we actually launched this thing upward. And, yet, we don't have any measure in here of how high it actually went. We don't have any measure in there of theta, and we don't have any measure in there of how long it was in the air. So, can that be right? Is it possible that this is all the information we need to calculate Vf? Don't you think we should have theta in there somewhere? What do you think? Let me ask you the following question to illustrate this. Let's say I take a ball, and I'm standing on top of a roof. And, I throw it straight up in the air with Vi. When it comes back down to where I am and I catch it, what is its speed? If I throw it up with Vi. Ignore air resistance. When it comes back down to me, how fast is it going? Same speed, right? Vi. If I throw it up with Vi, it's going to come back down with Vi. So, if I'm standing on a building, and I throw this thing straight up in the air, and now it comes back down past me. It's exactly the same as if I threw it down with speed Vi which tells you that the angle can't be involved there. And, this is kind of counterintuitive, right? But, this is really what happens. However, you throw that ball off the building, any direction you want. If you throw it with initial speed Vi, it's always going to hit the ground with a speed that is given by that equation right there. Square root of Vi squared plus 2gh. It's independent of which direction you throw it. Now, the final velocity will change direction, of course, but the speed will be exactly the same. Okay. Is that kind of weird? I think it's kind of weird. I've been doing physics a long time, and I still think that's kind of weird. Okay. You guys are more accepting than I am, I guess. Okay, so we solved the first part of this problem which was the second part of the problem which was what is the final speed. Let's go back to our problem again and see if we can solve for the time.