2. Vector Addition Recall that for vectors in one dimension (parallel vectors), the vectors are added algebraically. Vectors in 2 dimensions are add geometrically R Y θ X *X = (R) (cos θ ) *Y = (R) (sin θ )

3. Example 3A:Section Review Pg. 87 1, 2 2.) 126m 10 o = θ

4. R Y θ Vector Resultants with Right Triangles c 2 = a 2 + b 2 [R 2 = x 2 + y 2 ] θ R X 7 km h 6km/h R 2 = (6km/h) 2 + (7km/h) 2 R 2 = 36km 2 /h 2 + 49km 2 /h 2 R = 85 km 2 /h 2 R = ___________ km/h θ – theta: means the measure of the angle

5. We will not do this vector addition graphically. We will not do it using the Law of Cosines → We will do it trigonometrically or as the books says, analytically. All the Trig. You need to know for Physics I *SOH, CAH, TOA hypotenuse Side adjacent to θ θ Side opposite of θ Sin θ = Opp. Side Hypotenuse Cos θ = Adjacent side Hypotenuse Tan θ = Opposite side Adjacent side Solving Problems with 2 vectors in different directions:

6. Example Sample 3B page 93 *Add to your notes V r 2 = V x 2 + V y 2

7. Angles are in Standard Position What you learned in math is used in Physics! 0o0o 270 o 90 0 180 0 θ 50km/h 90 km/h Example : ?

8. Finding the resultant without R 2 = X 2 + Y 2 R = Resultant Tan θ = 50 km/hr =.5556 90 km/hr θ = 29.05 o Cos 29.05 o = 90km/hr R R = 90 km/hr.8742 R = 102.9 km/hr R = 102.9 km/hr @ 29.05 o STD position I II

9. Motion in 2 dimensions I. Projectiles: Objects thrown or projected into the air are projectiles whose parabolic paths are called trajectories. Motion of a projectile along a trajectory is called projectile motion. Max Height *V x = (V i ) (cos θ ) *V y = (V i ) (sin θ ) Once a projectile loses contact with the hand, bat, or gun barrel, it is only accelerated by g, in the X direction, a = 0 θ ViVi VxVx VyVy

10. Motion in 2 dimensions Max Height *V x = (V i ) (cos θ ) *V y = (V i ) (sin θ ) We can analyze the motion separately along each axis: X directionY direction X = (V x )(t) + (½)at 2 = (V x )(t) X = (V x )(t) = (Vi cos θ )t θ Y X

11. Motion in 2 dimensions Max Height *V x = (V i ) (cos θ ) *V y = (V i ) (sin θ ) We can analyze the motion separately along each axis: X directionY direction X = (V x )(t) + (½)at 2 = (V x )(t) θ Y X

12. Motion in 2 dimensions Max Height *V x = (V i ) (cos θ ) *V y = (V i ) (sin θ ) We can analyze the motion separately along each axis: X directionY direction X = (V x )(t) + (½)at 2 = (V x )(t) Y = (V y )(t) + ½gt 2 X = (V x )(t) = (Vi cos θ )t θ Y X

13. Motion in 2 dimensions Max Height *V x = (V i ) (cos θ ) *V y = (V i ) (sin θ ) We can analyze the motion separately along each axis: X directionY direction X = (V x )(t) + (½)at 2 = (V x )(t) Y = (V y )(t) + ½gt 2 X = (V x )(t) = (Vi cos θ )t Y = (V y )(t) + ½gt 2 = (V i sin θ )t + ½gt 2 * t = ½ time of flight for Y θ Y X

14. * Time of Flight Equation : (derived from the Y equation) Y = v i sin θ (t) + ½ gt 2 2v i sin θ = t g or 2v y = t g

15. Calculating X max without time : X max = 2 V i 2 sin θ cos θ g X max = 2 V i 2 sin 2 θ g

16. Example : A stone is thrown horizontally at 15m/s from the top of a cliff that is 44 m high. A. How long does the stone take to reach the bottom of the cliff? B. How far from the base of the cliff does the stone strike the ground? C. Sketch the trajectory of the stone.

17. Acceleration only vertically means: the time aloft is independent of the horizontal component of the trajectory. Max height = max Y Same t: time * if v i is the same Fly out Pop up Y

18. Homework #1 BK & WKBK A-C Chp. 3 (15) Book Sections 3A,3B& 3C 3A: Pg. 91 1-4 2.) 45.6m at 9.5 o East of North 4.) 1.8m at 49 o below the horizontal 3B: pg. 94 1,3,5 (answers back of the book) 3C: pg. 97 1,2,4 2.) 7.5km at 26 o above the horizontal 4.) 171 km at 34 o East of North WKBK 3A & 3B 3A: 3.) 7 jumps at 36 o West of North 5.) 65 o below the water 533m 3B 4.) x= 29m y=21m 5.) V x =335km/h V y =89.8km/h 6.) d= 900m x=450 m East y= 780m North

19. Workbook 3 D-E 3D 1. 5.98 m = X 3. 170m = Y 4. y= -.735m 3E 1. V i = 45.8 m/s 2. V=V x = 68.2 m/s 4. V i =20.8 m/s Y max = 11m Y max = 22.1m Homework #2 Chp. 3 Book & WKBK 3D-3E (11) BOOK-3D Problems pg. 102 2,3,4 2.) 4.9m/s 3.)7.6m/s 4.) 5.6m 3E problems pg. 1043 & 4 4.) 7m/s

20. *Scalars – has magnitude with units *Vectors – have magnitude with direction → Vectors can be added and subtracted creating a Resultant ( R ) 3N4N 7N 6N 3N