It's professor Dave, let's learn about projectile motion. We've learned how to use kinematic equations to describe the motion of an object moving horizontally, like a car, as well as objects moving vertically, like objects falling straight down to the ground, but what about objects that move in both of these directions? This type of motion, if it involves an object that is thrown or launched into the air, can be referred to as a projectile motion. Imagine a cannonball being fired at some angle from the horizontal. It will travel some distance up into the air before eventually falling back down and hitting the ground, some distance away from the cannon, and we can use a parabola to represent the path of this object. The important thing to understand about these kinds of examples is that the horizontal motion and vertical motion of the cannonball are completely independent of one another. This means we can use separate equations to discuss the motion in each direction, one equation that exclusively corresponds to the x-coordinates of the object and another that exclusively corresponds to the y-coordinates of the object. To drive this idea home consider two marbles, one dropped from a particular height and another that rolls off of a surface at that same height with some horizontal velocity. If these begin falling at the same time they will strike the ground at the same instant because their vertical motion is independent of any horizontal motion. The one with horizontal velocity will cover some distance in the X direction but it will fall downward at the same rate as the one that falls straight down and so they will have identical airtimes. How can we apply this to other real-world examples? well we have been throwing lots of rocks off of cliffs lately so let's throw one more. This time let's throw a rock at an upward angle of 30 degrees off the horizontal from the very edge of our favorite 100-meter cliff and with an initial velocity of 8.5 meters per second. Again we will ask how long before the rock hits the ground? and now additionally we want to know how far away from the edge of the cliff it will land. First things first let's make sure we understand how these questions relate to motion in both the x and y direction. The time it spends in the air only relates to Y direction behavior because it will stop being in the air when it hits the ground, no matter what the horizontal velocity is, from 0 to some huge number. The vertical motion will be independent of that. The distance it travels from the edge of the cliff depends on the horizontal velocity but also the time it spends in the air because once it hits the ground it can't travel any further. And the velocity at any moment can also be split into components. The horizontal velocity will be the same at every moment in this trajectory as long as we disregard wind resistance, but the vertical velocity will be the greatest at the moment the rock is thrown and then decrease until it reaches zero at the zenith and then become increasingly negative until it hits the ground. This is because there is a constant acceleration in the negative direction due to gravity. We know we can use these equations to answer these questions but in order to discuss the x and y directions separately we have to split up this velocity vector into x and y components. This is pretty simple to do. We can simply draw horizontal and vertical vectors starting from the same point as this one and extending precisely as far as this one does in the X or Y direction. These can be labeled V sub X and V sub Y and we can think of these as being the legs of the right triangle with the existing vector as the hypotenuse. To find out the magnitude of these vectors we just do a little trig. We know that the cosine of 30 will be equal to the adjacent leg over the hypotenuse which means 8.5 cosine 30 or 7.36 will be the initial velocity in the X direction. We also know that the sine of 30 will be equal to the opposite leg over the hypotenuse so 8.5 sign 30 or 4.25 will be the initial velocity in the y-direction. These velocities are independent of one another, meaning that a rock that was thrown straight up with an initial velocity 4.25 m/s will land on the ground at precisely the same time as the one thrown at 8.5 meters per second but at this 30 degree angle. That means that if we want to figure out how long it is in the air this is no different than a problem with one-dimensional vertical motion provided that we understand that we are only looking at the y component of this velocity vector. So we can plug the numbers in, -100 for the displacement, 4.25 for initial velocity and our usual acceleration. Putting it into standard form and plugging into the quadratic equation we will take the positive result for t and say that the rock will be in the air for 4.97 seconds. This also makes it easy to calculate how far the rock will travel, because we know the horizontal velocity will be 7.36 m/s times the 4.97 seconds that it is in the air which leaves us about 36.6 meters away from the edge of the cliff So projectile motion is a little more complicated than one-dimensional motion but if we know how to divide any vector into x and y components we can easily analyze projectile motion as the combination of two types of one-dimensional motion. Let's check comprehension. Thanks for watching, guys. Subscribe to my channel for more tutorials, support me on patreon so I can keep making content, and as always feel free to email me: