1. Projectile Motion Falling things, and rockets ‘n’ that… AP Physics Unit 1 Oct 10 Lesson 2

2.  Homework Oct 10 – Glencoe Page 164 – 165 #’s 45 52, 57, 59  Homework Oct 14– Serway  Page: 101-103  #’s 5&6(v), 9, 15, 17, 20-21  Lab Set up

3. Objectives : Do Now : How far above the ground will a ball above the ground, if it is thrown from ground level @ 30.656 m/s straight up in 2.50 seconds? Use 9.81 m/s 2 for gravity Y = Vi – ½ gt 2 Y = 30.625 – ½ (9.81)(2.5) 2 :

4. Objects Under the Influence of Gravity:

5. Continued… What does this mean for objects dropped from the same height? Objects dropped from the same height will hit the ground at the same time, regardless of weight (ignoring differences in air resistance) http://www.animations.physics.unsw.edu.au/jw/projectiles.htm#1 Solve the same way you would any other kinematics problem… AICR State what you know, what do you want to find? Choose an equation and solve for your unknown. Look out for key words like.. “dropped”, “from rest” and “stationary:”

6. Solving Problems Involving AICR Projectile Motion 1. Read the problem carefully, and choose the object(s) you are going to analyze. 2. Draw a diagram. 3. Choose an origin and a coordinate system. 4. Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g. 5. Examine the x and y motions separately. 6. List known and unknown quantities. Remember that v x never changes, and that v y = 0 at the highest point. 7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.

7. Projectile Motion Projectile motion follows a parabolic path i.e. http://phet.colorado.edu/sims/projectile-motion/projectile-motion_en.html It has a horizontal (x) component and a vertical (y) component. The horizontal component is not affected by gravity, so use separate equations for the different components.

8. Projectile Motion Equations Horizontal Component: X = Vx(t) Vx = Vi COS  Vertical Component: Y = Vy(t) –(1/2)gt 2 Vy = Vi SIN  t = 2 Vy /g (if Y = 0) t = √{2y/g} (if Vy =0)

9. Projectile Motion A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

10. It can be understood by analyzing the horizontal and vertical motions separately. Initial Vertical Velocity = ZERO t = √{2y/g} (if Vy =0) Initial Horizontal Velocity = Zero Projectile Motion

11. The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g. This photograph shows two balls that start to fall at the same time. The one on the right has an initial speed in the x-direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.

12. Projectile Motion If an object is launched at an initial angle of θ 0 with the horizontal, the analysis is similar except that the initial velocity has a vertical component. Initial Vertical Velocity does NOT = ZEROt = 2 Vy /g (if Y = 0) Initial Horizontal Velocity does NOT = Zero

13. Projectile Motion Day Two Oct 13 Objectives: Gain proficiency in collecting Data to solve for 2-D motion variables : Do NOW: How far down range will a cannonball go if fired at an angle of 30.0 deg an initial velocity of 18.5 m/s? assume no air resistance.

14. What will hit the ground first a bullet dropped or a bullet fired from the same height? Discuss with your neighbor… http://www.youtube.com/watch?v=D9wQVIEdKh8 http://www.youtube.com/watch?v=D9wQVIEdKh8

15. Projectile Motion Day Two Problems Projectile Motion Water slide Water Slide 2 Giant House Slide

16. EXAMPLE 4: DRIVING OFF A CLIFF!! How fast must the motorcycle leave the cliff to land at x = 90 m, y = -50 m? v x0 = ? y is positive upward y 0 = 0 at top v y0 = 0 v x = v x0 = ? v y = -gt x = v x0 t y = - (½)gt 2 Time to Bottom: t = √2y/(-g) = 3.19 s v x0 = (x/t) = 28.2 m/s

17. v x0 = v 0 cos(θ 0 ) = 16.0 15.97 m/s v y0 = v 0 sin(θ 0 ) = 12.0 36 m/s y = y 0 + v y0 t – (½)g t 2 t= 2.53 1 sec (quadratic formula) x = v x0 tx = 40.5 40.49 m t/2 = 1.27 sec Ymax = y 0 + v y0 t – (½)g t 2 = 1+ 12.0 (1.27 ) – 4.9 (1.27 ) 2 Y max = 8.39 26 m v 0 = 20.0 m/s, θ 0 = 37.0º Hang Time:2.53 seconds Range: 40.5 m Max Height: 8.39 m

18. ProjectileProjectile Motion Day Three Lab Lab You have probably watched a ball roll off a table and strike the floor. What determines where it will land? Could you predict where it will land? In this experiment, you will roll a ball down a ramp and determine the ball’s velocity with measure and watches. You will use this information and your knowledge of physics to predict where the ball will land when it hits the floor.

19. Test Review Unit 1 Motion in One-Dimension Displacement / Instantaneous Velocity / Acceleration Motion Diagrams Free Falling Objects / 1-Dimensional with constant Acceleration Motion in Two-Dimensions Vectors Coordinate System / Properties of Vectors Components of Vectors / Unit Vectors Adding Subtracting Multiplying Vectors (Graphically and Analytically) Scalar Products - Dot Products – (Honors: Cross Products) Motion in Two-Dimensions Projectile Range / Velocity Components / Maximum Height Time of Flight Circular Motion