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Projectile Motion

Projectile Motion. Half and Full projectiles. Definition of a projectile.

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Projectile Motion

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  1. Projectile Motion Half and Full projectiles

  2. Definition of a projectile Projectile motion is a form of motion where a particle (called a projectile) is thrown obliquely near the earth's surface, it moves along a curved path under the action of gravity. The path followed by a projectile is called its trajectory.

  3. Trajectory of a projectile The trajectory of a projectile is a parabola.

  4. Half projectile or horizontally launched projectile.

  5. Half projectile VX – Horizontal component remains constant. VY – Vertical component increases with descent just as free fall. (Marble drop and shot demo)

  6. How to solve for ½ projectile Because x-component remains constant, we use constant velocity equation to calculate Range. Vx= dx/t Here Vx - is the horizontal velocity. dx - is the range. t - is the time in of the fall. Range

  7. How to solve for ½ projectile A plane is moving At 100m/s. Suppose the plane drops a bomb every two second. How far apart are they going to land Use Vx= dx/t dx= Vx t = 200m Range

  8. Y – component increases due to gravity Remember the dropped and shot marble demo? Both take the same time. So we will solve the vertical part as a free fall. (same altitude and same time, so must be same.) We will use Galileo’s equations of motion in order to solve for the y-component. Since we are assuming that the vertical component is like a free fall, we can assume that Vi = 0m/s (initial velocity) a = -10m/s2 (acceleration due to gravity)

  9. Y – component increases due to gravity Since the vertical component is like a free fall, we can assume that: Vi = 0m/s (initial velocity) a = -10m/s2 (acceleration due to gravity) And apply to Galileo’s equations of motion: Vfy = Viy + at Vfy2 = Viy2 + 2ady dy=viyt+(½)at2

  10. Example Problem You and your buddy are skiing downhill and make a stunning jump. You CAN prove how fast you were going! If the ledge was 40 meters high and you travelled 52 meters horizontally in the air, how fast can you say you were going off the ledge? Step1: Solve for time using dy=viyt+(½)at2 t = 2.8s Step2: Solve for Vx using time from above. Vx= dx/t or Vx= 18.6m/s X Y dX = 52m dY = 40m Vi = 0m/s a = -10m/s2

  11. Review Question # 1 A stone is thrown horizontally from a cliff. If horizontal speed is doubled: a) what happens to the range? ANS: Double the range. b) What happens to the time of flight? ANS: Stays the same. c) What happens to the vertical velocity just before it hits the ground? ANS: Stays the same. d) What happens to the final velocity with which its hits the ground? ANS: Increases; V2 = (2VX)2 + (VY)2

  12. Review Question # 2 A stone is thrown horizontally off a cliff. If it is dropped with the same speed, but from a cliff twice as high; a) What happens to the Horizontal component of velocity? What happens to the range? ANS: VX remains the same, but range increases. b) What happens to the time of flight? ANS: Time of flight increases. c) What happens to the vertical velocity just before it hits the ground? ANS: It also increases. d) What happens to the final velocity with which its hits the ground? ANS: It also increases

  13. Full Projectiles The Maximum range of a projectile is at 45º. A projectile reaches maximum altitude when launched vertically. Range of a 30° is the same as the range of a 60° projectile. A satellite is a good example of a projectile that continually falls around the earth. The range of a full projectile is double that of the half projectile.

  14. Full Projectiles A projectile reaches maximum altitude (height) when launched vertically. If angle of launch is less than 45°, then the x-component (=Vcosθ) is greater than its y-component (=Vsinθ) of velocity.

  15. How to find the launch velocity of a projectile Step 1: Launch a projectile straight up (90°). Step 2: Find the time in air. Step 3: Plug into Galileo’s Equation and find Viy Vfy = Viy + at (Use half that time, a=-10m/s2, and Vfy=0) Viy= V the launch velocity.

  16. Known launch velocity and angle of launch. If you know the launch velocity and the angle of the launch. Step 1:Find Vx = V(Cosθ) Vy = V(Sinθ)

  17. Known launch velocity and angle of launch. Step 2: Use Vfy = Viy + at to find ½ the time of flight. (Viy =0 VY found in step 1 is the Vfy) Step 3: Use Vx from step1 and time from step2 to calculate the range, Dx. Dx

  18. Known launch velocity and angle of launch. Step 4: Using dy=viyt+(½)at2 find maximum height of the projectile. (Viy =0) dY

  19. How to solve for full projectiles If the launch velocity and the angle of launch is given. Step 1: Find Vx = Vcosθ and Vy = Vsinθ Now solve it as you would have solved for a ½ projectile. Use Vy = Vfy and solve. Why is Vy = Vfy? In a free fall Vi=Vf. So Viy = Vfy for a full projectile. If we solve for only the 2nd half of the projectile, then we can treat Vy=Vfy

  20. How to solve for full projectiles If the Range and the maximum height is given, then Step 1: Using maximum height (Dy) and the equations of the ½ projectile, find the time of flight. Step 2: Use range and time to find Vx. Step3: Use Dy to find Vfy (Which is Vy). Step 4: Use Vx and Vy to find V (Using Pythagorean Theorem) and θ using trig functions.

  21. How to solve for full projectiles If the angle of launch and the maximum height is given, then Step1: Use Dy to find Vfy (Which is Vy). Step 2: Use Vy = Vsinθ to find V. Now you know the launch velocity and the launch angle and you can solve it as previously explained.

  22. Horizontal graphs (DT and VT) X (m) V (m/s) T (s) T (s) A (m/s2) T (s)

  23. Vertical graphs for - (DT and VT) V(m/s) Y (m) 40 4 T (s) A (m/s2) Slope of the VT graph gives acceleration due to gravity = 10m/s2 on earth. T (s) 0 T (s) -10

  24. Cheddar Cheese V=5.0m/s 40° Neither half nor full projectile To solve such problems: Step1: Use given range to find the time it will take to reach that horizontal position. (Vx=Dx/t) Step2: Use that time to find the height reached in that time. (dy=Vit+(½)at2) Here (Vi=Vy) If dy> shown height, then he will make it. 0.35m 1.9m

  25. Cheddar Cheese V=5.0m/s 40° Neither half nor full projectile To find his velocity when he reaches the table, you have to find Vx and Vy and use Pythagorean theorem to find his actual velocity, V. Use trigonometric functions to find his angle of approach. 0.35m 1.9m

  26. Neither half nor full projectile You did the same calculations as shown in the two slides of Ralph S Mouse problem with the hoop lab problem.

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