 Hello, my name is Brad Langdill. I'm here to talk to you about projectile motion. I call this projectile motion type, too It's the second type that I talk about in class. It looks kind of like this. So imagine you've got a Canon and it's at some sort of angle and it's going to be fired in the air kind of like this shooting that cannonball up and Comes back down makes kind of a nice little parabola like so And this is a great animation from the University of Colorado the FET animation that shows us super well We'll take off the air resistance here We're going to try that again and what I want to look for is those velocity vectors those arrows so watch these arrows carefully here The arrow in the y direction is changing size, but the arrow in the x direction is not so we'll watch a little bit slower The x direction here has uniform motion the velocity is not changing the y direction has accelerated motion It's changing all over the place and we're going to use that idea as we go and solve this problem So here's an example of a type of problem that we might try. I've got a potato gun It's going to fire at five meters per second at 60 degrees. What's the range of the potato? Which means how far does it go horizontally? First thing you want to do draw to diagram I've got my velocity of five meters per second as my initial velocity and it's a diagonal vector at 60 degrees up in the air Then I've drawn in the x and y components of the velocity notice those components are tip-to-tail I've got the range the displacement in the x direction and I've got a final velocity Which is exactly the same as the initial velocity same triangle. It's just the directions are all well mostly They're all backwards the velocity is at a 60-degree angle going down The the y velocity is going downwards instead of upwards the x velocity is the same though Again uniform motion in the x direction. It doesn't change and I also put in the height if you need to calculate that and Different problem. We're not doing that here. Okay diagram done next thing. We're going to do We got to break that five meter per second velocity into x and y components We won't actually ever use five meters per second in any physics formula in this problem We use the x and the y components so I use trig to get my y component in my x component like you can see there Nothing too fancy. I recorded the decimal places down to four digits. This one worked out to exactly 2.5 So there's no more digits after that And I color-coded them as well So I made sure that I didn't accidentally mix up an x and a y later on when I did some calculations Alright first calculation I do I'm going to get the time get that out of the y dimension So I'm going to use an accelerated motion formula because in that animation We saw how the y direction had accelerated motion I'm using just the regular old acceleration formula acceleration is final velocity minus initial velocity divided by time I Put in the acceleration due to gravity. I actually typed it in as nine point one instead of nine point eight one But there we go. That looks better now The final velocity is negative Four point three three zero one now that's negative because again from the diagram. You can see it's going downwards The initial velocity is positive four point three three zero one Because those are both negative well the first one's a negative the second one's a positive when I subtract them It doesn't give zero it's going to give you know double that amount six eight point six six zero two meters per second Go through solve that for time and getting about point eight eight two eight seconds Now I made that purple because time isn't really an x component or a y component as a scalar not a vector So I didn't want to make it blue like the y components and I didn't want to make it red like the x Okay, now we know how long that object was in the air for in total next and last step I'm going to figure out how far it goes horizontally now in the x direction the horizontal direction It's going to be uniform motion. So I'm using just v equals d over t and notice I'm not using the five meters per second. I'm using the two point five meters per second vector I got to keep my x with my x so they want to figure out how far horizontally the projectile goes I have to use the velocity in the horizontal or x direction The time that I solved from the last part of the problem. I'm going to plunk in there I'm going to solve this and I get two point two meters So I think the key to this one is really to remember the x and y dimensions A lot of students have trouble because they will accidentally start off here by not thinking about breaking that five meters per second down So I hope that helps if you're looking for more problems and practice with this idea You can check out the website at ldindustries.ca