Announcements:

1. Announcements: • Physics Learning Resource Center Open, • > room P207-C • Open 9am - 5 pm Monday - Friday • Hours also listed on syllabus • Homework #1 (due this Fri. 9/8 by 5:00 pm EST on WebAssign) • Homework #2 (due next Fri. 9/15 by 5.00 pm)

2. Physics 151: Lecture 4Today’s Agenda 3-D Kinematics : • Review motion vs time graphs • Kinematics in 2 or 3 dimensions • Independence of x and y components • Projectile motion, baseball example

3. x t v vav • A few other useful formulas : t a t Review of 1-D Motion : • For constant acceleration we found:

4. Lecture 4, ACT 12-D Motion • Alice and Bill are playing air hockey on a table with no bumpers at the ends. Alice scores a goal and the puck goes flying off the end of the table. Which diagram best describes the path of the puck ? Alice Bill C) B) A)

5. We have already seen the 1-D kinematics equations. See text: 4-1 3-D Kinematics (Chapter #4) • The position, velocity, and acceleration of a particle in 3 dimensions can be expressed as: r = x i + y j + z k v = vx i + vy j + vz k(i,j,kunit vectors ) a = ax i + ay j + az k

6. See text: 4-2 and 4-3 3-D Kinematics • For 3-D, we simply apply the 1-D equations to each of the component equations. • Which can be combined into the vector equations: r = r(t) v = dr / dt a = d2r / dt2

7. See text: 4-4 3-D Kinematics • So for constant acceleration we can integrate to get: • a = const • v = v0 + a t • r = r0 + v0 t + 1/2 a t2 (where a, v, v0, r, r0, are all vectors)

8. See text: 4-5 2-D Kinematics • Most 3-D problems can be reduced to 2-D problems when acceleration is constant; • Choose y axis to be along direction of acceleration. • Choose x axis to be along the “other” direction of motion. • Example: Throwing a baseball (neglecting air resistance). • Acceleration is constant (gravity). • Choose y axis up: ay = -g. • Choose x axis along the ground in the direction of the throw.

9. y motion: a = -g y Reference frame on the ground. Reference frame on the moving train. x motion: x = v0t “x” and “y” components of motion are independent ! • A man on a train tosses a ball straight up in the air. • View this from two reference frames:

10. Projectile Motion. • If I set something moving near the earth, it reduces to a 2d problem we call projectile motion. • Use a coordinate system with x along the ground, y vertical with respect to the ground. (Notice no change in third direction.) • Equations of motion reduce to: • X: Dx = voxtax = 0 • Y: y = yo + voyt – g t2y positive upwards

11. Problem 1: • Sammy Sosa clobbers a fastball toward center-field. You are checking out your new fancy radar gun which can detect ball velocity, i.e. speed and direction. You measure that the ball comes off the bat with initial velocity is 36.5 m/s at an angle of 30o above horizontal. Since Sammy was hitting a high fastball, you estimate that he contacted the ball about one meter off of the ground. You know the dimensions of Wrigley field and the center-field wall is 371 feet (113m) from the plate and is 10 feet (3m) high. You decide to demonstrate your superfast math and physics skills by predicting whether Sammy get a home run before the play is decided.

12. v h  yo D Problem 1: • We need to find how high the ball is at a distance of 113m away from where it starts. Animation

13. y v h  x D Problem 1: • This is a problem in projectile motion. Choose y axis up. Choose x axis along the ground in the direction of the hit. Choose the origin (0,0) to be at the plate. Say that the ball is hit at t = 0, x = xo = 0, y = yo = 1m

14. Problem 1 • Variables • vo = 36.5 m/s • yo = 1 m • h = 3 m • qo = 30º • D = 113 m • a = (0,ay)  ay = -g • t = unknown, • Yf – height of ball when x=113m, unknown, our target

15. y v v0y  y0 v0x g x Problem 1 • For projectile motion, • Equations of motion are: vx = v0xvy = v0y - g t x = vx t y = y0 + v0y t - 1/ 2 g t2 And, use geometry to find vox and voy Find v0x = |v| cos . and v0y = |v| sin .

16. Problem 1 • Solve the problem, • Numbers: • y(t) = (1.0 m) + (113 m)(tan 30) - (0.5)(9.8 m/s2)(113 m)2/(36.5 m/s cos 30)2 = (1.0 + 65.2 - 62.6) m = 3.6m • Think about the answer, • The units work out correctly for a height (m) • It seems reasonable for the ball to be a little over 3m high when it gets to the fence. • Answer: since the wall is 3m high, and the ball is 3.26m high when it gets there, Sammy gets a homer.

17. y P v0 h  x L P y h x v0 sin(q) Typical questions :(projectile motion; for given v0 and q) • What is the maximum height the ball reaches (h) ? • How long does it take to reach maximum height ? • Would the answers above be any different if the projectile was moving only along y-axis (1-D motion) with the initial velocity: v0 sin (q) ? ( A ) YES ( B ) NO ( C ) CAN’T TELL

18. y v0 h P  x L Typical questions :(projectile motion; for given v0 and q) • What is the range of the ball (L) ? • How long does it take for ball to reach final point (P) ?

19. Lecture 4, ACT 2Motion in 2D • Two footballs are thrown from the same point on a flat field. Both are thrown at an angle of 30o above the horizontal. Ball 2 has twice the initial speed of ball 1. If ball 1 is caught a distance D1 from the thrower, how far away from the thrower D2 will the receiver of ball 2 be when he catches it ?(a) D2 = D1 (b) D2 = 2D1 (c) D2 = 4D1

20. Would the projectile miss or hit the target ? ( A ) MISS ( B ) HIT ( C ) CAN’T TELL t = 0 PROJECTILE t = 0 TARGET v0 x y Problem 2 (correlated motion of 2 objects in 3-D) • Suppose a projectile is aimed at a target at rest placed at the same height. At the time that the projectile leaves the cannon the target is released from rest and starts falling toward ground. t = t1

21. t = 0 TARGET y v0  t = 0 x PROJECTILE Problem 3 (correlated motion of 2 objects in 3-D) • Suppose a projectile is aimed at a target at rest somewhere above the ground as shown in Fig. below. At the same time that the projectile leaves the cannon the target falls toward ground. Would the projectile now miss or hit the target ? t = t1

22. t = 0 TARGET y h t = t v0  t = 0 PROJECTILE x D Solution (Problem 3) TARGET: y-component yT = y0 + v0y t - 1/ 2 g t2 yT = h - 1/ 2 g t2 PROJECTILE: y-component yP = y0 + v0y t - 1/ 2 g t2 yP = v0 sin(q) t - 1/ 2 g t2 • Prove that the projectile will hit the target yP = yT when (t) xP = D ! if P hits T : v0 sin(q) t =h but: t = D / v0x = D / v0 cos(q) so: v0 sin(q) / v0 cos(q) = h/D tan(q) = h / D • But this nothing else than the condition that the projectile is aimed at the target !

23. Problem from previous Exam-1 Two balls, projected at different times so they don’t collide, have trajectories A and B, as shown. Which statement is true. • Initial speed of ball B must be greater than that of ball A. • Ball A is in the air for a longer time than ball B. • Ball B is in the air for a longer time than ball A. • Ball B has a greater acceleration than ball A. • Ball A has a greater acceleration than ball B.

24. Recap of today’s lecture • Kinematics in 2 and 3 dimensions, Chapter 4.1-3 • Reading : • Chapter 4: Sections 4-5 • Solutions of Homework #1: • Will be available on the web: www.phys.uconn.edu/~dutta/151_2006 • To registration for webassign to http://www.webassign.net : • ID: first initial + last name (James S. Clark => jclark) • Institution:UConn • Password:your PeopleSoft ID (last 6 digits, no first 0 !) • let me know if you have problems. • Homework#2 (due next Fri. 9/15 by 5.00 pm