ï»¿ >> Hello class Professor Anderson here. Let's take a look at an example problem that might be rather relevant to you when you're working on your homework and this is the following problem: You're going to fire a bullet horizontally, ok, perfectly horizontally when you launch it and it's going to fly 50 meters to a target and it will fall 3 cm on its way. Ok so you are firing your rifle 50 meters to the target and the bullet's going to fall 3 cm on its way. So let's draw what this looks like. Here you are you've got your bullet that is launched horizontally and it's just going to fall a little bit. Ok so if we continue that horizontal line it would look like that. So this distance here we can call x final is 50 meters and this distance here is 3 cm. Ok let's ask the following question: Let's ask the question what is the speed of the bullet and how long does it take? What is the time equal to? Alright, we have a picture, we have some givens, we need to figure out how we're going to deal with this. So first off we need to draw a coordinate system and why don't we say that the coordinate system is right here this is x, this is y and if that is x and y then we can say x initial is 0, x final is 50 meters, y initial is 0, y final is what? What is y final equal to? Yeah >> Negative 3 centimeters. >> Negative 3 centimeters so if we do that in SI units it would be 0.03 meters yeah. What else do we know about the motion of this bullet? Well, it's projectile so a projectile has 0 acceleration in the x direction. Negative 9.8 in the y direction and we also know something else namely we said that the bullet is fired horizontally and if it's fired horizontally what is Vy initial? What does Vy initial have to be if it's fired perfectly horizontally? >> 0. >> 0 right, that's 0. Ok so this is everything we know. I mean it's quite a bit right? We know a lot of stuff so now let's grab some kinematic equations and see if we can answer these two questions, what's the speed of the bullet and how long does it take? So we want to figure out what the speed of the bullet is and we said that it's fired horizontally so Vy initial is 0 and so really all we're looking for is Vx initial. Alright, we might be able to do that because let's go to our kinematic equations. One of the kinematic equations we have is the following: X final equals X initial plus Vx initial times t plus 1/2 a sub x t squared. Alright, that looks pretty good. Let's plug in what we know. Let's plug in what we know. We know x final. We know x initial is 0. Vx initial is what we're looking for. We know that a sub x is 0 and so we get a nice little equation here x, f equals Vxi times t. Do we have enough in this equation to solve for Vxi? No we don't right? Xf we know but Vxi we don't know and t we don't know. So we need something else to help us solve for t and that's where the y motion comes into it. Y final equals y initial plus Vy initial times t plus 1/2 a sub y t squared and let's see what we know. We know y final right that is our negative 3 centimeters. Y initial we said was 0. Vy initial we said was also 0 since it was fired horizontally and ay is negative g and so now we have a nice little equation here that we can solve for t. Let's do that we'll solve it for t and we'll plug it back into our other equation. Yf equals minus 1/2 gt squared let's see if we can solve this thing for t. We have to multiply across by 2, it's actually a negative 2 right so we have negative 2 times Yf over g and we're going to take the square root of that and you're a little concerned at this point because there's a negative sign in there but we remember that Yf is also a negative number. So this becomes negative 2 times negative 0.03 meters we're all on SI units and we're going to divide that whole thing by 9.8. So plug that into your calculator and tell me what you get. We've got the square root of .06 over 9.8. We'll approximate that, that is 6 times 10 to the minus 2 over 10. We're going to take the square root of that and so that becomes 6 times 10 to the minus 3 but I don't know how to do that so we'll change it to 60 times 10 to the minus 4 and the square root of 60 is about 8, it's a little bit less than 8 we'll say it's 7.9 and then we have a 10 to the minus 2 and so we're going to say that this is 0.079 seconds, that's my guess. Am I anywhere close? What did you get? >> .078. >> .078 that one right there I'd say we were pretty close right? Would you agree? 0.078 seconds is how long that bullet is in the air. Ok so we've solved for t and now we want to solve for v. What is the speed coming off? Vxi is equal to x sub f over t. X sub f we know is 50 and we are now dividing by 0.078. Tell me what you guys get for that. [ pause in speaking ] Anybody get an answer for that? 50 over 0.078. Yeah what did you get? >> 641 >> 641 meters per second. Ok, which is pretty fast right? This is our sort of typical speeds of bullets but that's moving right that's very fast. Remember the rule for going from meters per second to miles per hour you roughly double it ok so that's 1200 miles per hour or something like that. Ok any questions about this one how we approach this? Ok good, hopefully that's clear. If not, definitely come see me in office hours.