Projectile Motion

Projectile Motion

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 ydirection. – Air resistance is ignored in projectile motion unless explicitly stated. – The path of a projectile is parabolic in nature.

The Trajectory of a Projectile

Choosing Coordinates & Strategy For projectile motion: – Choose the y-axis for vertical motion where gravity is a factor (ay g -9.81 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.

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 -9.81 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. ix ix i ix iy ix ix iy v f vi

Formulas You Will Use x-direction: – dx vix t vfxt (velocity is constant since ax 0) – vix vi cos y-direction: – dy ½ (vi vf) t vavg t – vfy viy gt – dy viy t ½ g(t)2 – vfy2 viy2 2gd – viy vi sin

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 ax 0

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.] 1. Determine the horizontal and vertical components of the ball’s initial velocity. 2. Determine the maximum height reached by the ball. 3. Determine the total amount of time the ball is in the air. 4. Determine how far the ball travels before it hits the ground.

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. vi . 8 / m 5 35 vix s viy

Example #1 (Part 1 - cont.): Using trigonometry (see reference table). – viy vi sin (8.5 m/s)(sin 35 ) 4.9 m/s – vix vi cos (8.5 m/s)(cos 35 ) 7.0 m/s v i 8. /s m 5 35 vix viy

Example #1 (Part 2): Finding the Maximum Height Whenever a projectile motion question (1) vf vi at asks about anything related to motion in the vertical direction, you should (2) d vit ½ at2 automatically know that you will need a (3) v2 vi2 2ad 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 -9.81 m/s2. List the variables that you know and don’t know viy vfy ay g dy t 4.9 m/s 0 m/s -9.81 m/s2 ? ? 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

Example #1 (Part 3): Finding the time of flight As before, list all the known and unknown variables. (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 ? 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

Example #1 (Part 3): An Alternative Way Instead of using equation (1) (1) vf vi at we will use equation (2) this time. (2) d vit ½ at2 We will also analyze the problem (3) v2 vi2 2ad 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: t 2d y 2 1.2m 0.49s 2 g 9.81m / s 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.

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: (1) vf vi at dx vixt (2) d vit ½ at2 (3) v2 vi2 2ad 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

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.] 1. Sketch the theoretical path of the projectile. 2. Calculate the horizontal distance, d, if the projectile’s total time of flight is 2.5 seconds. 3. Determine the height, h, of the platform. 4. Determine the final velocity, vf, when the projectile hits the ground.

Example #2(Part 1): Sketching the path. All projectiles followed a curved or parabolic path.

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

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. viy vfy ay g dy t 0 m/s ? -9.81 m/s2 ? 2.5 s 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. (1) vf vi at (2) d vit ½ at2 (3) v2 vi2 2ad dy viyt ½ gt2 dy (0 m/s)(2.5 s) – ½ (9.81m/s2)(2.5 s)2 dy 31 m

Example #2(Part 4): Determining the final velocity. Start by listing all variables known. vix vfx viy vfy ay g dy t 30 m/s 0 m/s ? -9.81 m/s2 31 m 2.5 s 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) (1) vf vi at vfy viyt gt 2 (2) d v t ½ at 2 i vfy (0 m/s)(2.5 s) – (9.81m/s )(2.5 s) (3) v2 vi2 2ad vfy -24.5 m/s

Example #2(Part 4): Determining the final velocity. 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 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. vfx Applying the Pythagorean Theorem, we get: vf2 vfx2 vfy2 v f v 2fx v 2fy v f (30m / s ) 2 (24.5m / s ) 2 v f 38.7 m / s vfy vf ?