How To Get Horizontal And Vertical Components

By | February 19, 2017

Higher Jumps Vertical Jump Training Resistance Bands

Speaker 1: Today we've got a bit of a jumpchallenge. I'm going to show you a little bit of a workout here. It's going to add alittle bit of an endurance portion of it as well as vertical leap. Grab you a nice boxthat's a little challenging for yourself, get your bands on and then lay out a benchor anything that you're going to jump over, it's going to be a barrier. So make it a littlebit challenging, but make sure that you can get over and back. Let me show you real quickand then we'll run through a set. So I got my box height here, make sure you got yourbands on. Let me just tell you when you're wearing the bands and you're working on yourvertical leap you want to make sure that you

control your knees. Don't let them controlyou. A lot of benefits, we've talked about this in hundreds of other tutorials I've postedfor you, but you got to make sure that when you're wearing the bands when you land workagainst the resistance and you're going to build your hips so quick, but if you let themcontrol you or you're not strong enough to do these types of things without the bandson, you probably want to start doing these types of exercises without the bands first.So let's jump into it here. Get your box height a little bit challenging. We're going to goless reps, but I'm going to show you both exercises. Get a good arm sweep on top ofthe box. Soft landings, you either hop back

or just step off, whatever you feel comfortablewith. The most important thing is you get up, get down, get up, get down. It's not afast pace, we're going for height. Nice soft landings. The second part of the exercisewould be speed, but under control. We're going to go over, all about getting your knees up.Let's jump into the workout. We're going to go 8 max jump heights and gofor a 15 second burst right after that for speed. Let's go and get started. 8 reps righthere. That's one, two, three. Make sure that you're landing neutral like that. Don't letyour feet come together, stay under control. I think I got two more. The last one. good. Now we got about 15 secondburst here for speed. Try to get up, get around,

get back over. Readyé Good. Make sure youget your knees up on those. If you start getting dizzy, I got a little bit dizzy there at theend, take your time, maybe don't spin around so fast. Catch your breath, that's a littlebit of endurance. Run through that, three or four sets and push yourself hard.

Demonstrating the Components of Projectile Motion

Mr.P.: Good morning. Today we're going todemonstrate the components of projectile motion by building velocity and acceleration vector diagrams. Bo: Hey guys. Billy: Hey Bo. Bobby: Hi Bo. ♫ (lyrics) Flipping Physics ♫

Mr. P.: Let's start with a quickreview of projectile motion. Bobby, please remind me, what is true of the motion of a projectilein the XDirection. Bobby: In the XDirection, the projectile will move at a Constant Velocity. Mr. P.: Correct, in the Xdirection the velocity of a projectile is constant. Billy, what about the motion

of a projectile in the YDirectioné Billy: In the YDirection, a projectile experiencesUniformly Accelerated Motion with an acceleration equal to the negative of the acceleration due to gravity. On Earth, that accelerationwould be equal to 9.81 meters per second squared. Mr. P.: Yes, in the Ydirectionthe acceleration is constant

therefore we can use theUAM equations of motion. Therefore, projectile motion is made up of a horizontal and a vertical component. The Horizontal Componentis Constant Velocity. And the Vertical Component isUniformly Accelerated Motion. With a downward accelerationthat has a magnitude of 9.81 meters per second squared. Now, let's combine them all.

Again, notice how the Xdirectionhas a Constant Velocity. And the Ydirection has auniform downward acceleration. Now let's add velocity vectorcomponents in the Xdirection. Bobby, what should thevelocity vector components look like the Xdirection for the ball as it moves to the righté Bobby: Well, in the Xdirection the ball has a constantvelocity to the right,

so the vector shouldhave the same magnitude the whole time and bepointed to the right. Mr. P.: Correct. So, as the ball moves to the right we can freeze the ballat several locations, and add the velocityvector in the Xdirection which will always have the same magnitude and therefore the same length arrow.

Math 142 75 Part 1 Magnitude Direction Horizontal and Vertical Components of Position Vectors

In part one of our lesson on algebraically defined vectors you will learn how to find the magnitude, direction horizontal and vertical components of position vectors so first let's understand what position vector notation is

the vector u is equal to sharp bracket a,b sharp bracket and what that means is this is the vector that has an initial point of the origin and has a terminal point a,b so remember, when there's parentheses around the a,b that names a single point

when there's these pointy brackets around the a,b it means the vector that starts at the origin and ends at the point a,b now the magnitude we know is the length of vector u, and that is indicated by this symbol it looks like absolute value of u, but what it really is is magnitude of vector u and the magnitude of u

is acting just like r when we're using our Trig Definitions and we know that r is equal to the square root of x squared plus y squared so then that means that the magnitude of u is going to equal the square root of a squared plus b squared you want to make sure that you understand that when we write vector u then that's the name of the vector

that's like calling someone Joe or Susie whereas, when we write magnitude of u then that is the length of the vector so that will be a number. That would be like saying 6 feet tall so vector u is like using someone's proper name magnitude of u is a number representing the length

of the vector the direction of vector u is given by theta and applying our Trig Definitions we know then that the tangent of theta would equal b over a, because it's like always, the y value divided by the x value the convention with vectors is that theta will be in the interval 0° to 360 degrees

Leave a Reply