Motion in Two Dimensions

1. Motionin Two Dimensions Chapter 7.2

2. Projectile Motion • What is the path of a projectile as it moves through the air? • Parabolic? • Straight up and down? • Yes, both are possible. • What forces act on projectiles? • Only gravity, which acts only in the negative y-direction. • Air resistance is ignored in projectile motion.

3. Choosing Coordinates & Strategy • For projectile motion: • Choose the y-axis for vertical motion where gravity is a factor. • Choose the x-axis for horizontal motion. Since there are no forces acting in this direction (of course we will neglect friction due to air resistance), the speed will be constant (a = 0). • Analyze motion along the y-axis separate from the x-axis. • If you solve for time in one direction, you automatically solve for time in the other direction.

4. Fg The Trajectory of a Projectile • What does the free-body diagram look like for force?

5. ay The Vectors of Projectile Motion • What vectors exist in projectile motion? • Velocity in both the x and y directions. • Acceleration in the y direction only. vx (constant) ax = 0 vy (Increasing) Trajectory or Path • Why is the velocity constant in the x-direction? • No force acting on it. • Why does the velocity increase in the y-direction? • Gravity.

6. Formulas for Motion of Objectsassuming d is displacement from origin and time starts at 0.

7. There are only so many things a projectile can do Go up and down Be shot at an angle Fall d = vi t + ½ at2 vf = vi + at vf 2=vi 2+2ad Beware cases where t <> hang time Solve x and y separately. t is when the object hits the ground. d = ½ at2 vf = vi + at vf 2=vi 2+2ad Note: vi<>0 is possible Move horizontally Be shot horizontally Be shot at an angle d = vave t vf = vi + at vave = ½(vf+vi) vf 2=vi 2+2ad Solve x and y separately. t is when the object hits the ground. Solve x and y separately. t is when the object stops.

8. Ex. 1: Launching a Projectile Horizontally • A cannonball is shot horizontally off a cliff with an initial velocity of 30 m/s. If the height of the cliff is 50 m: • How far from the base of the cliff does the cannonball hit the ground? • With what speed does the cannonball hit the ground? • Note: This is the same problem as “a life raft is dropped from an airplane …”

9. a = -g 50m vx vf = ? vy x = ? Diagram the problem vi Fg = Fnet

10. State the Known & Unknown • Known: • xi = 0 • vix = 30 m/s • yi = 0 • viy = 0 m/s • a = -g • y = -50 m • Unknown: • x at y = -50 m • vf = ?

11. Perform Calculations (y) • y-direction: • vy= -gt • y = viyt – ½ gt2 • Using the first formula above: • vy = (-9.8 m/s2)(3.2 s) = 31 m/s

12.   Perform Calculations (x) • x-Direction • x = vixt • x = (30 m/s)(3.2 s) = 96 m from the base. • Using the Pythagorean Theorem: • v = vx2 + vy2 • v = (30 m/s)2 + (31 m/s)2 = 43 m/s

13. Ex. 2: Projectile Motion above the Horizontal • A ball is thrown from the top of the Science Wing with a velocity of 15 m/s at an angle of 50 degrees above the horizontal. • What are the x and y components of the initial velocity? • What is the ball’s maximum height? • If the height of the Science wing is 12 m, where will the ball land?

14. y x vi = 15 m/s viy a = -g  = 50° vi = 15 m/s vix  = 50° 12 m Ground x = ? Diagram the problem Fg = Fnet

15. State the Known & Unknown • Known: • xi = • yi = • vi = •  = • a = • Unknown: • ymax = ? • t = ? • x = ? • viy = ? • vix = ?

16. State the Known & Unknown Known: xi = 0 yi = 12 m vi = 15 m/s  = 50° a = -g Unknown: ymax = ? t = ? x = ? viy = ? vix = ?

17.  = 50° vi = 15 m/s vyi vxi Perform the Calculations (ymax) • y-direction: • Initial velocity: viy = visin • viy = (15 m/s)(sin 50°) • viy = 11.5 m/s • Time when vfy = 0 m/s: vfy = viy + at = viy – gt • 0 = viy – gt  viy = gt  t = viy / g • t = (11.5 m/s)/(9.81 m/s2) • t = 1.17 s • Determine the maximum height: d = vit + ½ at2 • dmax = viyt – ½ gt2, plus 12 m • ymax = 12 m + (11.5 m/s)(1.17 s) – ½ (9.81 m/s2)(1.17 s)2 • ymax = 18.7 m

18.   Perform the Calculations (t) • Since the ball will accelerate due to gravity over the distance it is falling back to the ground, the time for this segment can be determined as follows • Time when ball hits the ground: ymax = viyt – ½ gt2 • Since yi can be set to zero as can viy, • t = 2*ymax/g • t = 2(18.7 m)/ (9.81 m/s2) • t = 1.95 s • By adding the time it takes the ball to reach its maximum height to the time it it takes to reach the ground will give you the total time. • ttotal = 1.17 s + 1.95 s = 3.12 s

19.  = 50° vi = 15 m/s vyi vxi Perform the Calculations (x) • x-direction: • Initial velocity: vix = vicos • vix = (15 m/s)(cos 50°) • vix = 9.64 m/s • Determine the total distance: x = vixt • x = (9.64 m/s)(3.12 s) • x = 30.1 m

20. Analyzing Motion in the x and y directions independently. • x-direction: • dx = vix t = vfxt • vix = vicos • y-direction: • dy = ½ (vi + vf) t = vavg t • vf = viy + gt • dy = viy t + ½ g(t)2 • vfy2 = viy2 + 2gd • viy = visin

21. Key Ideas • Projectile Motion: • Gravity is the only force acting on a projectile. • Choose a coordinate axis that where the x-direction is along the horizontal and the y-direction is vertical. • Solve the x and y components separately. • If time is found for one dimension, it is also known for the other dimension.