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Projectile Motion

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**A Projectile What is a projectile?**

A projectile is any object that is acted upon by gravity alone. Note that gravity acts in the negative y-direction. Air resistance is ignored in projectile motion unless explicitly stated. The path of a projectile is parabolic in nature.

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**The Trajectory of a Projectile**

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**Choosing Coordinates & Strategy**

For projectile motion: Choose the y-axis for vertical motion where gravity is a factor (ay = g = m/s2). Choose the x-axis for horizontal motion. Since there are no forces acting in this direction, the speed will be constant (ax = 0). Analyze motion along the y-axis separate, or independently from motion along the x-axis. This is a step that most students have difficulty with. Note that time is the only variable that will always be the same for both the vertical and horizontal directions. Hence, if you can find if for the y-direction, you also have it for the x-direction, and vice-versa.

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Strategies Continued If the projectile is fired horizontally, then viy will be zero. If the projectile is launched at an angle greater than 0o, then vy will be 0 m/s at the very peak of its trajectory. A common misconception is that the acceleration = 0 m/s2 at the peak as well. But it actually equals m/s2, and anywhere else along the path of the projectile The magnitude of the initial launch velocity will be the same as the final velocity if the projectile lands at the same height from which it is launched. iy ix i vf = vi

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**Formulas You Will Use x-direction: y-direction:**

dx = vix t = vfxt (velocity is constant since ax = 0) vix = vicos y-direction: dy = ½ (vi + vf) t = vavg t vfy = viy + gt dy = viy t + ½ g(t)2 vfy2 = viy2 + 2gd viy = visin

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**The Vectors of Projectile Motion**

What vectors exist in projectile motion? Velocity in both the x and y directions. Note that it increases in the y-direction while it is constant in the x-direction. Acceleration in the y direction only. vx (constant) vy (Increasing) ay = 9.81 m/s2 Trajectory or Path ax = 0

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**Example #1: Determining the vertical and horizontal components, and maximum height of a projectile**

A child kicks a soccer ball with an initial velocity of 8.5 meters per second at an angle of 35º with the horizontal, as shown. The ball has an initial vertical velocity of 8.5 meters per second at 35 relative to the horizontal. [Neglect air resistance.] Determine the horizontal and vertical components of the ball’s initial velocity. Determine the maximum height reached by the ball. Determine the total amount of time the ball is in the air. Determine how far the ball travels before it hits the ground.

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**Example #1 (Part 1): Determine the vertical and horizontal components of the initial velocity.**

For the vertical and horizontal velocities, you need to use component vector analysis using trigonometry. Begin by creating a right triangle with the vertical and horizontal components making up the a and b legs of the right triangle. 35 vi = 8.5 m/s viy vix

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**Example #1 (Part 1 - cont.): Using trigonometry (see reference table).**

viy = visin = (8.5 m/s)(sin 35) = 4.9 m/s vix = vicos = (8.5 m/s)(cos 35) = 7.0 m/s 35 vi = 8.5 m/s viy vix

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**(0 m/s)2 = (4.9 m/s)2 – (2)(9.81 m/s2)(d)**

Example #1 (Part 2): Finding the Maximum Height Whenever a projectile motion question asks about anything related to motion in the vertical direction, you should automatically know that you will need a formula involving the acceleration due to gravity. Substitute g wherever you see a. In this example, ask yourself what do you know about the ball’s motion at the top of its trajectory. You know that the vertical component of velocity at the peak of the soccer ball’s trajectory will be 0 m/s. Therefore, we will set vfy = 0 m/s. In the vertical direction, we know that gravity will act on the soccer ball such that it will cause it to accelerate at m/s2. List the variables that you know and don’t know Choose a formula from the three in the table above that contains all the known variables with only the unknown one missing. Hence you will use equation (3) from the table above: vfy2 = viy2 + 2gd (0 m/s)2 = (4.9 m/s)2 – (2)(9.81 m/s2)(d) Solving for d yields: d = 1.2 m (1) vf = vi + at (2) d = vit + ½ at2 (3) v2 = vi2 +2ad viy vfy ay = g dy t 4.9 m/s 0 m/s -9.81 m/s2 ?

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**(0 m/s) = (4.9 m/s) – (9.81 m/s2)(t)**

Example #1 (Part 3): Finding the time of flight As before, list all the known and unknown variables. You have two choices of equations that you can use, (1) or (2) Solving for t using equation (1): vfy = viy + at (0 m/s) = (4.9 m/s) – (9.81 m/s2)(t) t = 0.5 s However, this time only covers half the flight of the ball, therefore, the total time is double, or 1.0 s (1) vf = vi + at (2) d = vit + ½ at2 (3) v2 = vi2 +2ad viy vfy ay = g dy t 4.9 m/s 0 m/s -9.81 m/s2 1.2 m ?

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**Example #1 (Part 3): An Alternative Way**

Instead of using equation (1) we will use equation (2) this time. We will also analyze the problem starting from the second half of the trajectory. Since the velocity is 0 m/s at the top of its motion, viy = 0 m/s. Equation (2) then simplifies to: dy = ½ gt2 Solving for t gives us: As before, this time only accounts for ½ of the whole trajectory of the ball, therefore you must double it to get a total time of 1.0 s. (1) vf = vi + at (2) d = vit + ½ at2 (3) v2 = vi2 +2ad

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**Example #1 (Part 4): Finding the Horizontal Distance**

To find the horizontal distance, you only need to be concerned with motion in the x-direction. As previously mentioned, there is no acceleration in the x-direction, therefore, vix = vfx = vx = constant. Equation (2) can be used since the portion containing the acceleration (½ axt2) will be zero. Hence: dx = vixt Since the initial velocity in the x-direction and the time have already been determined: dx = (7.0 m/s)(1.0 s) = 7.0 m (1) vf = vi + at (2) d = vit + ½ at2 (3) v2 = vi2 +2ad

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Example #2: Determining the horizontal distance, height and final velocity of a projectile launched horizontally. A projectile is launched horizontally at a speed of 30. meters per second from a platform located a vertical distance h above the ground. The projectile strikes the ground after time t at horizontal distance d from the base of the platform. [Neglect friction.] Sketch the theoretical path of the projectile. Calculate the horizontal distance, d, if the projectile’s total time of flight is 2.5 seconds. Determine the height, h, of the platform. Determine the final velocity, vf, when the projectile hits the ground.

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**Example #2(Part 1): Sketching the path.**

All projectiles followed a curved or parabolic path.

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**Example #2(Part 2): Determining the Horizontal Distance.**

As in the previous example, the velocity in the horizontal direction is constant. dx = vixt dx = (30 m/s)(2.5 s) = 75 m

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**dy = (0 m/s)(2.5 s) – ½ (9.81m/s2)(2.5 s)2**

Example #2(Part 3): Determining the height of the platform. As in the previous example, we will list those variables we know and those that we do not. Since we want to find dy, we will have to use either equation (2) or (3). Equation (2) is a better fit since we do not know the final velocity yet. dy = viyt + ½ gt2 dy = (0 m/s)(2.5 s) – ½ (9.81m/s2)(2.5 s)2 dy = 31 m viy vfy ay = g dy t 0 m/s ? -9.81 m/s2 2.5 s (1) vf = vi + at (2) d = vit + ½ at2 (3) v2 = vi2 +2ad

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**Example #2(Part 4): Determining the final velocity.**

Start by listing all variables known. The final velocity, vf, is the sum of both the vertical and horizontal components of the final velocity, where we will use the Pythagorean Theorem. But first, we need to find the final velocity in the vertical direction using equation (1) vfy = viyt + gt vfy = (0 m/s)(2.5 s) – (9.81m/s2)(2.5 s) vfy = m/s vix = vfx viy vfy ay = g dy t 30 m/s 0 m/s ? -9.81 m/s2 31 m 2.5 s (1) vf = vi + at (2) d = vit + ½ at2 (3) v2 = vi2 +2ad

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**Example #2(Part 4): Determining the final velocity.**

A vector diagram can help to fully understand how the final velocity is the sum of the vector components in the horizontal and vertical directions. Applying the Pythagorean Theorem, we get: vf2 = vfx2 + vfy2 vix = vfx viy vfy ay = g dy t 30 m/s 0 m/s -24.5 m/s -9.81 m/s2 31 m 2.5 s vf = ? vfx vfy

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2D Motion 2015 cjcb Angled Projectile Motion. RECAP: What is Projectile Motion? Projectile motion is the motion of an object that is being thrown or launched.

2D Motion 2015 cjcb Angled Projectile Motion. RECAP: What is Projectile Motion? Projectile motion is the motion of an object that is being thrown or launched.

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