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Mechanics Lecture 2, Slide 1 Vectors and 2d-Kinematics Continued Relevant Equations How to use them Homework Hints.

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Presentation on theme: "Mechanics Lecture 2, Slide 1 Vectors and 2d-Kinematics Continued Relevant Equations How to use them Homework Hints."— Presentation transcript:

1 Mechanics Lecture 2, Slide 1 Vectors and 2d-Kinematics Continued Relevant Equations How to use them Homework Hints

2 Hyperphysics-Trajectories Mechanics Lecture 1, Slide 2

3 Projectile Motion Quantities Mechanics Lecture 2, Slide 3 Initial velocity  speed,angle Maximum Height of trajectory, h=y max Range of trajectory, D Height of trajectory at arbitrary x,t “Hang Time” Time of Flight, t f

4 Hyperphysics-Trajectories Mechanics Lecture 1, Slide 4

5 Maximum Height of Trajectory Mechanics Lecture 2, Slide 5 Height of trajectory,h=y max

6 Time of Flight Mechanics Lecture 1, Slide 6

7 Time of Flight, “Hang Time” Mechanics Lecture 2, Slide 7

8 Hyperphysics-Trajectories Mechanics Lecture 1, Slide 8

9 Range of trajectory Mechanics Lecture 2, Slide 9

10 Angle for Maximum Range Mechanics Lecture 2, Slide 10 MAXIMUM range OCCURS AT 45 0

11 Will it clear the fence Mechanics Lecture 1, Slide 11

12 Height of Trajectory at time t or position x Mechanics Lecture 2, Slide 12 Height of trajectory, y(x) Height of trajectory, y(t)

13 Projectile Trajectory Equations Mechanics Lecture 1, Slide 13 Height of trajectory as f(x), y(x) Height of trajectory as f(t), y(t) Range of trajectory Time of Flight (“Hang Time”) Maximum height

14 Where will it land? Mechanics Lecture 1, Slide 14

15 Launch Velocity-Given R and  Mechanics Lecture 1, Slide 15

16 Launch Angle Mechanics Lecture 1, Slide 16

17 Launch Velocity –Given R and h Mechanics Lecture 1, Slide 17

18 Mechanics Lecture 2, Slide 18 Field Goal Example A field goal kicker can kick the ball 30 m/s at an angle of 30 degrees w.r.t. the ground. If the crossbar of the goal post is 3m off the ground, from how far away can he kick a field goal? y-direction v oy = v o sin(30 o ) = 15 m/s y = y o + v oy t + ½ at 2 3 m = 0 m + (15 m/s) t – ½ (9.8 m/s 2 ) t 2 t = 2.8 s or t = 0.22 s. x-direction v ox = v o cos(30 o ) = 26 m/s D = x o + v ox t + ½ at 2 = 0 m + (26 m/s)(2.8 s) + 0 m/s 2 (2.8 s ) 2 = 72.8 m D 3 m y x Illini Kicks 70 yard Field Goal

19 Homework Hints-Baseball Mechanics Lecture 1, Slide 19

20 Homework Hints- Baseball Stadium Wall Mechanics Lecture 1, Slide 20

21 Homework Hints – Stadium Wall Mechanics Lecture 1, Slide 21 Calculate time to reach wall using v x : Calculate y position at time to reach wall:

22 Homework Hints-Catch Mechanics Lecture 1, Slide 22

23 Homework Hints-Catch Mechanics Lecture 1, Slide 23

24 Homework Hints-Catch Mechanics Lecture 1, Slide 24

25 Homework Hints-Catch 2 Mechanics Lecture 1, Slide 25

26 Homework Hints-Catch 2 Mechanics Lecture 1, Slide 26 v V x is constant ! Kinetic energy should be same as when ball was thrown. Y- component of velocity would be downward.

27 Homework Hints-Catch 2 Mechanics Lecture 1, Slide 27 Same conditions as before

28 Homework Hints – Soccer Kick & Cannonball Mechanics Lecture 1, Slide 28

29 Homework Hints – Soccer Kick & Cannonball Mechanics Lecture 1, Slide 29

30 Homework Hints – Soccer Kick & Cannonball Mechanics Lecture 1, Slide 30

31 Trigonometric Identity for range equation Mechanics Lecture 2, Slide 31

32 Trigonometric Identities relating sum and products Mechanics Lecture 2, Slide 32 List of trigonometric identities

33 Question 2 Mechanics Lecture 2, Slide 33

34 Question 2 Mechanics Lecture 2, Slide 34

35 Hyperphysics-Trajectories Mechanics Lecture 1, Slide 35


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