 Hello, my name is Brad Langdell. I'm here to talk to you about projectile motion. I call this projectile motion type 1. It's the first time I type I talk about in class, and it looks kind of like this. So we've got a cannon. It's going to fire horizontally, and it's some distance up in the air. It looks kind of like that. That went pretty fast. Let's try that again. Let's watch it a little bit slower. Here's what I want you to look at. Check out those arrows. These are velocity vectors. Now the thing I want you to notice about these velocity vectors, as that cannon ball goes through the air, let's try it again. Is that this blue vector never changes in length. It's always the same. It's the x velocity, and the red vector, it's the y velocity, increases as you go. That's because there is accelerated motion in the y direction due to the acceleration due to gravity and the gravitational force, but in the x direction there's uniform motion. The object doesn't speed up, it doesn't slow down. We're going to keep that in mind as we solve this problem. So we've got an object thrown horizontally with a velocity of 4 meters per second from the top of a 15 meter tall building. What's the range of the object? It's a fancy way of saying how far does it go horizontally. Okay, first thing to do in a problem like this, you want to draw a diagram. So pretty similar to what we just saw in that cannon example, and I've color coded the vectors in the x dimension and the y dimension. So in the x dimension I know how fast it's going horizontally. That's my velocity in the x. I'm trying to figure out displacement in the x direction. I know the velocity in the y direction and I know the acceleration due to gravity in the y direction. And one thing that I like to do is I like to put a negative sign on that displacement in the y direction because after all that object is going to go downwards. Alright, so I've got my diagram set up. Now I'm going to start thinking a little bit about accelerated and uniform motion. So I'm going to start off here in the y direction. Keeping my y vectors with my other y vectors. So I'm going to use a kinematics formula that has the displacement in the y direction. Again, that's a negative number because we're going to go, the ball's going to fall downwards. The acceleration due to gravity, another y dimension variable. The initial velocity in the y direction and time. Now here's one of the cool keys to this problem. You might be thinking, hey isn't the velocity 4 meters per second? No, no, that's the x dimension velocity. This is the y dimension velocity. And it's zero because initially the cannonball wasn't moving in the y direction. So that makes the first term of this kinematics formula we're going to use, d equals vi t plus 1 half at squared, end up just being zero. That first term cancels right out. So here's where I go and substitute it into that formula. Put in the negative sign on the 15 meters. Put in the acceleration due to gravity. And I did my algebra to go through and solve for the time. I recorded that down to four decimal places and I made it purple because time isn't really an x or a y variable. It's a scalar and it's going to be applying in both dimensions. So drew a diagram, used a kinematics formula, in this case d equals vi t plus 1 half at squared to figure out how long it would take that projectile to hit the ground. This is the same calculation we did when we were just dropping a ball just vertically downwards from some height and we wanted to know how long it took to hit the ground. Exactly the same calculation. Okay, third and final step. Now we think about the x dimension. In the x dimension there's uniform motion, which means that we can just use the good old fashioned uniform motion formula, v equals d over t. The thing I'm going to be really careful with here is I'm going to use the velocity in the x direction in order to get a displacement in the x direction. Keep your x with your x. So four meters per second, plug in my time that I got from the last problem and I round it off to two significant digits. It works out to 7.0 meters. So that's a really standard type one projectile motion problem. If you're looking for more questions like this, you can check out my website, LDindustries.ca.