WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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In this lesson we are going to talk about projectile motion.
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Our objectives for the lesson include analyzing situations of projectile motion in uniform gravitational fields.
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Writing down expressions for both the horizontal and vertical components of velocity and positions as functions of time
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and using those expressions and analyzing the motion of a projectile that was projected with some arbitrary initial velocity.
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What is a projectile?
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A projectile is an object that is acted upon only by gravity.
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In reality we know that air resistance plays a role but as we began our study of projectile motion we are going to neglect air resistance.
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When we talk about projectiles typically what we are talking about is something that is launched at an angle
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and follows some sort of path something like that.
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Although technically something going straight up and down as a projectile that it is not really nearly as interesting.
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Let us look at the path of the projectile.
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Projectiles launched an angle move in parabolic arcs.
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Let us draw a baseline here.
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If we have some nice flat level ground and we launch some object to some initial velocity let us call it Vi for V initial.
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At some angle θ i with respect to the ground it is going to follow a parabolic arc.
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It is going to come up and come back down in a symmetric path assuming we neglect air resistance.
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A couple of interesting things to point out here.
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However, the long it takes to go up is the exact same amount of time it takes to go down assuming it returns to level ground.
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Its initial speed right when it leaves the ground is its speed right before it hits the ground.
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Its initial angle is going to be the same angle it makes with the ground over here so we have this symmetry of motion.
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We could look at the maximum height that our projectile attains which is halfway through its path.
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There is our maximum height.
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We could also talk about the range of our projectile.
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How far it travels horizontally?
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Our range would be that distance.
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Projectiles follow parabolic paths and assuming they are coming down to the same level they left at it will follow it will have a symmetric path.
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As we study these, one thing that will help us simplify this tremendously is looking at the independence of motion.
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When projectiles are launched at an angle they have some piece of vertical motion and some piece of horizontal motion.
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We can treat those 2 separately.
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The vertical motion is acted upon by the acceleration due to gravity by the force of gravity.
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Horizontal motion there is no acceleration because there is no air resistance.
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As we split those up we can greatly simplify our problem solving process.
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Let us take a look with an example.
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Fred throws a baseball 42 m/s horizontally from a height of 2m.
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How far will the ball travel before it reaches the ground?
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Let us start off with a little diagram.
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He throws it from a height of 2m.
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It is the same as if we tossed a ball off a cliff of a height of 2m, not that much of a cliff.
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Here is our ball and it is traveling horizontally at 42 m/s.
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Our height here is 2m.
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Let us take a look at what we know.
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Horizontally we know the velocity the speed is 42 m/s and that is going to remain the same because there is no acceleration until it is acted upon by some other force.
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Once it hits the ground at which point or into a different studying that projectile motion any longer.
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We will have some change in horizontal position Δ x and it will take some amount of time to do this.
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We could also look at the vertical components of its motion.
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Vertically its initial velocity is 0.
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It is moving horizontally but up and down vertically initially it is not moving.
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Let us call down our positive y direction since that is the direction it moves first and the vertical.
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We will have some final velocity we do not know that yet.
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It will travel some vertical distance Δ y which will be 2m, +2m because we call down the positive y direction.
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It will accelerate in the y direction at 9.8 m/s² g+ because we call down positive.
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We do not yet know the time.
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As we look at these what is the same between horizontal and vertical motion the amount of time the ball is in the air.
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The time is the one thing that will be the same between our 2 tables of information.
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If we can find the time it is in the air vertically we could then use that to figure out how far it went horizontally.
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Let us do that.
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Let us see if we can find a way with what we know to solve for the time vertically.
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I think I would start with our kinematic equation Δ y = V initial t + 1/2 ay t².
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It is nice here since V initial is 0 that whole term become 0 and we can simplify this to Δ y = 1/2 at² ay t²
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or t² is going to be equal to 2 Δ y / ay which will be 2 × 2m / 9.8 m/s².
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If I want just t I take the square root of that so if I want t I will take the square root of that and find that t comes out to about 0.639s
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that is how long it takes for the ball to hit the ground to travel the 2m vertically.
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That is the same as the amount of time it is traveling horizontally so that 0.639 s over there.
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Since we have a constant velocity horizontally we can go right to our equation to find how far it travels.
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Δ x = the velocity in the x direction × time over which it is traveling that speed or 0.639 s × velocity 42 m/s.
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I come up with a horizontal distance that travels of 26.8m.
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A little bit more involved but really it is just breaking up a problem into horizontal and vertical components to keep things nice and simple mathematically.
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Let us take a look at another example.
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The diagram represents the path of a stunt car that is driven off a cliff neglecting friction.
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Compared to the horizontal component of the car's velocity over here at a, the horizontal component of the car's velocity at b
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is we need to realize here the horizontal component of velocity is not going to change because acceleration due to gravity is down.
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It is not affecting the horizontal motion of the car.
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Therefore, we can say the horizontal component of the velocity is the same.
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If it asks about the vertical component of the velocity well of course it is beating up the longer it falls.
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For objects launched at an angle in order to figure out the components of velocity we typically
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have to break that velocity up to x and y components using their trigonometry.
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We will use those initial velocity components in our horizontal and vertical motion tables.
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An object will travel the maximum horizontal distance on level ground with the launch angle of 45° as long as you are neglecting air resistance.
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That comes up quite a bit.
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If you want an object to go the furthest possible distance what angle do you want is 45°.
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Let us take a look at what this would look like if we have something that is launched with a velocity at an angle.
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There is our ground and let us say we have some initial velocity vector right there.
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We will call that V initial.
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It is launched at some angle θ with respect to the ground.
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In order to find its components we will look at the x component first.
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That piece we call the initial x and the component vertically V initial in the y.
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From our trigonometry remember V initial y is just going to be V initial sin θ and V initial x the x component of that velocity vector
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is going to be V initial cos θ because we are at the adjacent side of that angle.
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Sin θ because we are the opposite side here.
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For range, if we want the maximum possible range on our projectile.
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If we want to go as far as possible, there is our range.
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The trick is a launch angle of 45° assuming it is on level ground and you are neglecting air resistance of course.
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Let us take an example of the projectile launched in an angle.
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Herman, the human cannonball is launched from level ground at an angle of 30° above the horizontal with the initial velocity of 26 m/s.
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Interesting career path for Herman.
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How far does he travel horizontally before reuniting with the ground?
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Now let us start by taking a look at his initial velocity vector.
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I am going to make some axis here.
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These initially launched an angle of 30° above the ground, above the horizontal and that initial velocity is 26 m/s.
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The x component the horizontal component of that velocity we can find the initial x is just going to be 26 m/s cos 30° or 22.5 m/s.
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The y component the initial y will be 26 m/s sin 30° or about 13 exactly 13 m/s sin 30 is ½.
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How far did he travel horizontally before reuniting with the ground?
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Again, I think the most important thing for us to know here is how long he is in the air.
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We will start by answering that with a vertical analysis.
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Vertically we have V initial, V final, some horizontal displacement, ay and t.
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Herman's path is going to take him up and down like that.
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Remember our trick for simplifying this is let us look at either the way up or the way down that by cutting this in half.
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Let us say that we want to look at the way up.
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Time to go up, time to go down, and V initial then if we are going up we are going to call positive y up.
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V initial is 13 m/s because we said up was positive, V final at its highest point it stops.
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We do not know the Δ y.
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We know the acceleration is -9.8 m/s² negative because we call up positive and the acceleration is down due to gravity.
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We are trying to find the time and we can use V = V initial + at.
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We arrange that to find the time is V - V initial / a is 0 -13 m/s / -9.8 m/s².
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Our time is going to be about 1.33s but note that is the time just to go up.
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If it takes 1.33s to go up how long does it take to go up and down?
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Twice that so our total time is twice the time it takes to go up or 2.6 and if we do not round here I get closer to 2.65s.
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We know the time in the air and we can figure out the range how far Herman travels horizontally.
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Let us do that.
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Horizontally we do not know Δ x yet but we know the velocity in the x direction was our 22.5 m/s.
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We found that out initially by breaking at vector in the components.
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And the total time Herman's traveling horizontally is 2.65s.
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We just found that by analyzing the vertical motion.
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Δ x = the velocity in the x × the time which is 22.5 m/s × 2.65s for 59.6m.
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Let us take a look at some motion graphs then.
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An arrow was launched from level ground with initial velocity V0 at an angle of θ above the horizontal.
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Sketch a graph of the displacement velocity and acceleration of the arrow was functions of time and neglect air resistance.
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Let us do this for both x and the y so we are going to make some graphs here.
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Make another graph for velocity and another graph over here for our acceleration.
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This those a little bit but we will get the idea across I think and our x axis.
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We will have a look at the x up here and look at the y down here.
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Take a look at graphs of position vs. time, velocity in the x vs. time.
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Acceleration in the x vs. Time and for the y we will have Δ y for time and we will have the velocity
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and the y compared the time and will have acceleration in the y in time.
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Let us start off on the right hand side with acceleration usually those are an easy place to start.
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In the x direction we know there is no acceleration so that graph is nice and flat at 0 no acceleration.
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In the y direction the entire time our acceleration is 9.8 m/s² down.
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Assuming we are calling up positive let us just draw our line down here at -9.8 m/s² for acceleration.
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Now let us move backwards and take a look at the velocity.
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In the x direction the velocity must be constant.
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We have some constant velocity and that makes sense because if we take the slope of velocity which we will get our acceleration.
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The slope of a straight line is 0 so our acceleration is 0.
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Down here for the y we have a negative acceleration which means our velocity must have a negative slope.
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For something launched from level ground with initial velocity of V 0 at an angle it is going to start off that is maximum velocity.
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It is going to slow down stop and then start going faster and faster in the negative direction.
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There is our graph of velocity vs. time in the y direction.
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Note that the slope is negative in constant and our acceleration is negative in constant.
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Now for the displacement vs. time.
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We know that we are changing our position horizontally at a constant rate.
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We would see something that looks kind of like this.
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Assuming we are starting from the point we call 0.
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We are going to have a 0 nice straight linear line slightly redundant.
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The slope of this is positive in constant.
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I will look positive in constant velocity.
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All of these match up when we look at our slopes.
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When we look at displacement in the y direction what we are going to have change in values here
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because we have a constant but we have a change in value here so we have changing slopes.
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Right in the middle the value of our velocity is 0, right in the middle we must have a slope of 0.
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I would draw something kind of like this for our vertical displacement.
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We start off and 0 we reach a maximum here halfway through the slope here is 0 corresponding to a velocity of 0.
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Then we start coming back the other way steeper and steeper negative slope more and more negative velocity.
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That is how I would graph our displacement velocity and acceleration for each of these components for an object that is launched from the ground.
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Graphing projectile motion.
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Contract a vector component as a function of time or plot their path as a y = function of x.
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We have been doing is function of time but if we wanted to do this is as a function of x solve for the time in the horizontal
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and then we can use that to eliminate time in the vertical equation.
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What does that mean?
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Let us look for an analysis of something launched an angle horizontally Δ x is our initial x velocity × time.
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And therefore time would be our horizontal displacement divided by our horizontal velocity.
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When we move to our vertical analysis Δ y = Vy initial t + 1/2 ay t².
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We can now substitute in what we know for time from our horizontal equation recognizing that t = Δ x / our initial x velocity.
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Δ y = V initial y × our time Δ x / V x initial + ½ ay × Δ x² / x initial².
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Or just cleaning this up a little bit we could write this as our y position.
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Δ y assuming we are starting at 0 equals if we do this as a function of x we are going to have Vy initial / Vx initial ×
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our x coordinate + we will have ay / 2 Vx initial² × x².
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Y = some constant × x + another constant x² that is the form of a parabola, a parabolic motion.
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Let us take a look at an example here.
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We got an air fighter from a tower, if an arrow was fired from a 15m high tower with an initial velocity 50 m/s and angle of 30° above the horizontal,
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find the height of the projectile above the ground as a function of its horizontal displacement.
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Since our object goes up first let us call up our positive y direction.
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We can take a look and figure out our initial x velocity, Vx initial is going to be V cos θ, V cos 30° or 50 m/s cos 30° which will be 43.3 m/s.
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Doing the same thing for the y the initial velocity in the y will be V sin θ, V sin 30° or 50 sin 30 sig 30 is half so that is just going to be 25 m/s.
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Since we called up the positive direction our acceleration in the y is -g or -9.8 m/s².
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We can go back to our form for the motion of a projectile in parabolic form.
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We had y = V initial / Vx initial x + ay / 2 Vx initial² x².
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Substituting in our values that we just determined here y = 25 / 43.3 x + -9.8 / 2 × 43.3² x².
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Let us set our origin to 0, 15 because that is where we are starting.
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I get that y = our 15m we are starting with + 25 / 43.3, .577x - 9.8 / the quantity of 2 × √43.3 it is going to be 0.00261 x².
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And there is our formula for the parabola that shows our x and y motion.
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Let us actually check this and let us make a graph of that.
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I plotted a nice pretty one in a software package that shows this formula y is a function of x.
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Notice that we start here at 15 m are going up, coming back down and we are going to hit the ground right around there.
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It should be pretty reasonable to say 1234 looks like we are about 244 m or so that is how far
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that arrow would travel before it struck the ground when launch from that 50m high tower.
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Alright speaking of launching things from a height.
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Black Bear the pirate fires a cannonball from the deck of his ship an angle of 30° above the horizontal with an initial velocity of 300 m/s.
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How far does the cannonball travel before contacting the ocean waters if the ship's deck is 8m above the waterline?
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A quick diagram here if there is our ground we are starting from 8m above and doing something like that.
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We have got 8m as our initial height.
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We will let us say let us take a look at our horizontal analysis.
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Horizontally our initial velocity in the x direction is going to be 300 cos 30 or about 260 m/s.
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Our final velocity is the same.
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There is no acceleration horizontally so that is 260 m/s.
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Our acceleration in the x is 0.
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We do not know how far it goes horizontally that would be nice to know, that is what we are after.
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We do not know time yet.
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How are we going to figure out time?
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Now I would go to the vertical.
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Let us take a look at our vertical analysis to figure out some more information.
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Vertically let us call up the positive direction since a cannonball travels up vertically first.
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V initial in the y direction is going to be 300 m/s sin 30° or 150 m/s.
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Let us say we have the final y we do not know yet.
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Δ y while the entire trip if it starts here and it ends up 8m below its total displacement is going to be -8m.
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-8 because we call up positive it is going to travel much further vertical distance but its displacement will be -8.
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Acceleration in the y will be -9.8 m/s² negative because we called up the positive direction.
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Let us see what we can find out here.
00:27:01.200 --> 00:27:11.000
As I do this it looks like we could solve for Vy first if we try to find t initially we are going to come up with a quadratic.
00:27:11.000 --> 00:27:15.200
Let us go with the finding final velocity in the y first.
00:27:15.200 --> 00:27:28.400
Vy² = V initial y² + 2ay Δ y our kinematic equation complies that Vy² is going to be equal to our initial velocity
00:27:28.400 --> 00:27:38.400
in the vertical 150² + 2 × -9.8 × -8.
00:27:38.400 --> 00:27:56.300
Vy² = 22,657 m² / s² or if I take the square root of that Vy = + or - about 150.5 m/s.
00:27:56.300 --> 00:27:59.300
We are going to take some common sense here.
00:27:59.300 --> 00:28:03.500
Go here when it gets this point is it traveling up or down.
00:28:03.500 --> 00:28:19.700
It must be going down vertically because we called up positive so we must be negative value Vy is -150.5 m/s.
00:28:19.700 --> 00:28:26.700
Find the velocity in the y -150.5 m/s.
00:28:26.700 --> 00:28:32.100
From here we can figure out the time it took to do that.
00:28:32.100 --> 00:28:45.100
Using another kinematic equation let us go with if ay is Δ Vy / t then t is going to be Δ Vy / a.
00:28:45.100 --> 00:29:03.100
That is going to be our -150.5 m/s -150 m/s our initial velocity divided by acceleration -9.8 m/s² or about 30.7s.
00:29:03.100 --> 00:29:10.700
T= 30.7s and that is the same amount of time that the cannonball is going to be traveling horizontally.
00:29:10.700 --> 00:29:16.900
I can fill in my time here horizontally as 30.7s.
00:29:16.900 --> 00:29:22.500
We can take a look now at the horizontal displacement of the cannonball.
00:29:22.500 --> 00:29:44.400
Δ x is going to be the x velocity × time or 260 m/s the horizontal velocity our time of 30.7s or about 7,970 m.
00:29:44.400 --> 00:29:49.800
It is quite the path that is traveling.
00:29:49.800 --> 00:29:53.200
Taking a look at one last question here.
00:29:53.200 --> 00:30:02.400
Kevin kicks a football across level ground, if the ball follows the path shown what is the direction of the balls acceleration when it gets to point b?
00:30:02.400 --> 00:30:06.000
Fairly common question but a trick question.
00:30:06.000 --> 00:30:11.800
The entire time the ball is in the air with the only force acting on it is gravity.
00:30:11.800 --> 00:30:23.800
Gravity pulls down so the acceleration on the ball is a very pretty down, straight down there is the direction of our acceleration.
00:30:23.800 --> 00:30:30.200
Any point in the path even when vertically it stops at the highest point the acceleration is always straight down.
00:30:30.200 --> 00:30:33.200
Thank you for watching www.educator.com, we will see you again soon.
00:30:33.200 --> 00:30:34.000
Make it a great day everyone.