Professor Anderson

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>> Hello class. Professor Anderson here. Let's take a look at something that we talked about in class that is maybe a little disturbing, and it's this. If we shoot a bullet horizontally, very far away, it's going to eventually drop to the ground. What if I take that same bullet and where I'm standing I just drop it straight to the ground? Okay, here's our ground level. Which one is going to hit the ground first? Well, it turns out they hit at exactly the same time. Now, it's kind of hard to fire a bullet perfectly horizontally, of course, but let's say you can do it. Alright, you fire this thing perfectly horizontally, and it goes way far away. It goes like a football field away. Does it really take the same amount of time as if I dropped the bullet straight down? Well, maybe. Let's convince ourselves. These are both projectiles, right? As soon as they leave the gun, or my hand, they are both projectiles, and so they are governed by the projectile equations which are the kinematic equations, x final equals x initial, plus v x i, times t, plus one-half a x t squared. And, y final equals y initial, plus v y initial times t, plus one-half a sub y, t squared. Now, let's think about this x equation for a second. In the case of the bullet, we'll call this one number one, and this one number two. Horizontal bullet is number one. Dropped bullet is number two. In the case of bullet number one, what do we have? We have x final equals where it started, zero, plus v x i times t, plus zero. For bullet number two we have zero equals zero, plus zero, plus zero. Not a lot of information in that equation right there, right? Except, to confirm that zero does equal zero no matter how many zeros you add to it, so that's good. What about over here? For bullet number one y final is the ground. y initial is however high we started from. We can call that h. v y initial is zero, if we fire it horizontally, a y is, of course, g. What about bullet number two? Bullet number two y final, is zero. It's on the ground. y initial is h, v y initial is zero. We dropped it from rest out of our hand. a y is minus one-half g t squared. So, look, those two equations are exactly identical. Okay, the y equations for the fired bullet versus the dropped bullet are exactly the same, versus the dropped bullet are exactly the same, and so you can say, ah, t one has to equal t two, those equations are exactly the same. And in fact, you can solve one of them to get t. It's not that complicated. The difference is in the x equations. One of them travels very far in that amount of time, in the x direction. The other one doesn't travel anywhere in the x direction. Okay, so it's a little counterintuitive when you first think about it, but if you remember that these equations are decoupled, the x motion is decoupled from the y motion, then it makes sense. Okay, is that clear to everyone? Alright, good. If not, definitely you can see me in office hours. Cheers.

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>> Hello class. Professor Anderson here. Let's take a look at something that we talked about in class that is maybe a little disturbing, and it's this. If we shoot a bullet horizontally, very far away, it's going to eventually drop to the ground. What if I take that same bullet and where I'm standing I just drop it straight to the ground? Okay, here's our ground level. Which one is going to hit the ground first? Well, it turns out they hit at exactly the same time. Now, it's kind of hard to fire a bullet perfectly horizontally, of course, but let's say you can do it. Alright, you fire this thing perfectly horizontally, and it goes way far away. It goes like a football field away. Does it really take the same amount of time as if I dropped the bullet straight down? Well, maybe. Let's convince ourselves. These are both projectiles, right? As soon as they leave the gun, or my hand, they are both projectiles, and so they are governed by the projectile equations which are the kinematic equations, x final equals x initial, plus v x i, times t, plus one-half a x t squared. And, y final equals y initial, plus v y initial times t, plus one-half a sub y, t squared. Now, let's think about this x equation for a second. In the case of the bullet, we'll call this one number one, and this one number two. Horizontal bullet is number one. Dropped bullet is number two. In the case of bullet number one, what do we have? We have x final equals where it started, zero, plus v x i times t, plus zero. For bullet number two we have zero equals zero, plus zero, plus zero. Not a lot of information in that equation right there, right? Except, to confirm that zero does equal zero no matter how many zeros you add to it, so that's good. What about over here? For bullet number one y final is the ground. y initial is however high we started from. We can call that h. v y initial is zero, if we fire it horizontally, a y is, of course, g. What about bullet number two? Bullet number two y final, is zero. It's on the ground. y initial is h, v y initial is zero. We dropped it from rest out of our hand. a y is minus one-half g t squared. So, look, those two equations are exactly identical. Okay, the y equations for the fired bullet versus the dropped bullet are exactly the same, versus the dropped bullet are exactly the same, and so you can say, ah, t one has to equal t two, those equations are exactly the same. And in fact, you can solve one of them to get t. It's not that complicated. The difference is in the x equations. One of them travels very far in that amount of time, in the x direction. The other one doesn't travel anywhere in the x direction. Okay, so it's a little counterintuitive when you first think about it, but if you remember that these equations are decoupled, the x motion is decoupled from the y motion, then it makes sense. Okay, is that clear to everyone? Alright, good. If not, definitely you can see me in office hours. Cheers.