## Projectile motion part 1 Onedimensional motion Physics Khan Academy

Welcome back. I'm not going to do a bunch ofprojectile motion problems, and this is because I thinkyou learn more just seeing someone do it, and thinkingout loud, than all the formulas. I have a strange notion that Imight have done more harm than good by confusing you with a lotof what I did in the last

couple of tutorials, so hopefullyI can undo any damage if I have done any, or even betterhopefully, you did learn from those, and we'lljust add to the learning. Let's start with ageneral problem. Let's say that I'm at the topof a cliff, and I jump instead of throwing something,I just jump off the cliff. We won't worry about my motionfrom side to side, but just assume that I gostraight down.

We could even think that someonejust dropped me off of the top of the cliff. I know these are getting kindof morbid, but let's just assume that nothingbad happens to me. Let's say that at the topof the cliff, my initial velocity velocity initial isgoing to be 0, because I'm stationary before the persondrops me or before I jump. At the bottom of the cliffmy velocity is

100 meters per second. My question is, what is theheight of this cliffé I think this is a good timeto actually introduce the direction notion ofvelocity, to show you this scalar quantity. Let's assume up is positive,and down is negative. My velocity is actually 100meters per second down I

could have assumedthe opposite. The final velocity is 100 metersper second down, and since we're saying that downis negative, and gravity is always pulling you down, we'regoing to say that our acceleration is equal togravity, which is equal to minus 10 meters persecond squared. I just wrote that ahead oftimes, because when we're

dealing with anything ofthrowing or jumping or anything on this planet, wecould just use this constant the actual number is 9.81, butI want to be able to do this without a calculator, so I'lljust stick with minus 10 meters per second squared. It's pulling me down, so that'swhy the minus is there. My question is: I know myinitial velocity, I know my final velocity, right before Ihit the ground or right when I

### Projectile at an angle Twodimensional motion Physics Khan Academy

So I've got a rocket here. And this rocket is goingto launch a projectile, maybe it's a rock of some kind, with the velocity often meters per second. And the direction of that velocity is going to be be 30 degrees, 30 degrees upwards from the horizontal. Or the angle between thedirection of the launch

and horizontal is 30 degrees. And what we want tofigure out in this tutorial is how far does the rock travelé We want to figure out how, how far does it travelé Does it travelé And to simplify this problem, what we're gonna do is

we're gonna break downthis velocity vector into its vertical andhorizontal components. We're going to use a vertical component, so let me just draw it visually. So this velocity vector can be broken down into its vertical and its horizontal components. And its horizontal components.

So we're gonna get somevertical component, some amount of velocityin the upwards direction, and we can figure, we can use that to figure out how long will this rock stay in the air. Because it doesn't matter what its horizontal component is. Its vertical component is gonna determine

how quickly it decelerates due to gravity and then reaccelerated, and essentially how long it's going to be the air. And once we figure outhow long it's in the air, we can multiply it by, we can multiply it by the horizontal component of the velocity, and that will tell us how far it travels.

And, once again, the assumption that were making this tutorials is that air resistance is negligible. Obviously, if there wassignificant air resistance, this horizontal velocitywould not stay constant while it's traveling through the air. But we're going to assume that it does, that this does not change,