v v y = v sinθ Component Vectors:

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1 Component Vectors: Recall that in order to simplify vector calculations we change a complex vector into two simple horizontal (x) and vertical (y) vectors v v y = v sinθ v x = v cosθ 1

2 Component Vectors: (review) Recall that in order to simplify vector calculations we change a complex vector into two simple horizontal (x) and vertical (y) vectors v v y = v sinθ v x = v cosθ If v = 10 m/s [30 degrees up] what are v x and v y? Answer: v y = 5.0 m/s [up] vx = 8.7 m/s [right] 2

3 What type of motion does the ball have as it travels across the desktop? A. Uniform B. Accelerated 2.0 m/s 3

4 Uniform Motion 2.0 m/s 4

5 Uniform Motion 2.0 m/s Therefore the formula d x = v x x t applies 5

6 What type of motion would this ball have as it falls? A. Uniform B. Accelerated 6

7 Accelerated Motion 7

8 Accelerated Motion Therefore the following formulas apply: d y = v iy t + (a y t 2 )/2 [ no vf ] d y = v fy t (a y t 2 )/2 [no v i ] d y = ( v 2 fy v 2 iy ) / 2a y [ no t ] d y = [(v fy + v iy )/2] t [ no a ] a y = (v fy v iy ) / t [ no d ] a y = 9.80 m/s 2 8

9 How long does it take for this ball to hit the floor? 9

10 How long does it take for this ball to hit the floor? d y = v iy t + (a y t 2 ) / 2 gives t = 2d y / a y t = 0.45 s 10

11 Projectile Motion has both Uniform and Accelerated components to its motion. Uniform in the horizontal dimension and accelerated in the vertical dimension. Uniform motion Accelerated Motion 11

12 Projectile from a Plane Travelling Horizontally 12

13 How far does the ball travel to the right while it is falling? (ie What is the ball's range?) 2.0 m/s 13

14 How far does the ball travel to the right while it is falling? (ie What is the ball's range?) 2.0 m/s Answer: d x = v x x t d x = 2.0 m/s x 0.45 s = 0.90 m Read: Sec 3.2 pp Do : #s 1 3 p.86 14

15 What if the projectile were shot upward at an angle? 2.0 m/s 30 Pull Pull 15

16 First off, we'll consider a ball being kicked on a soccer field. 2.0 m/s 30 Pull The velocity at which it is kicked can be broken into vertical and horizontal components. Horizontally: v ix = 2 cos 30 = 1.7 m/s Pull Vertically: v iy = 2 sin 30 = /s 16

17 How long will the ball be in the air? Pull Time to rise: use a = (v f v i ) / t rearranging gives t up = (v fy v iy ) / a y v fy = 0 m/s, v iy = /s, a y = 9.80 m/s 2 t up = ( /s / 9.80 m/s 2 ) = 0.10 s Since the ball will land at the same level from which it is kicked, the time to fall will equal the time to rise. Therefore the total time in the air is 0.20 s 17

18 Another way to approach this problem is to use a second degree polynomial derived from: d y = v iy t + (a y t 2 )/2 or d y = v fy t (a y t 2 )/2 For this problem d y = 0 (since it lands at the same level from which it is kicked) ; v iy = /s and a y = 9.8 m/s 2 This gives 0 = t 4.9 t t = 1 so, t = 1 / 4.9 = 0.20 s 18

19 What is the ball's range? Range = d x = v x t = 1.7 m/s x 0.20 s = 0.34 m Try This: A ball is kicked at 5.0 m/s and an angle of 40. What is its time aloft and its range? HW: Complete #2a through d on page 94 19

20 What is the final velocity of the ball? The final velocity is made up of two components. There's a horizontal component v fx and a vertical component v fy. Recall v x is uniform therefore: v fx = v ix = v sin θ In our example v fx = 1.7 m/s v fy can be calculated using: d y = ( v fy 2 v iy 2 ) / 2a y 20

21 d y = ( v 2 fy v 2 iy ) / 2a y Rearranging gives: v fy = v 2 iy + 2a y d y (d y = 0) v fy = = /s [Down] = /s[up] This was probably a bit obvious for this problem since one would expect it to have the same vertical speed as when it started. We now have v fx = 1.7 m/s [R] and v fy = /s [Dn] We still need to find v f 21

22 We can solve for v f using the Pythagorean Theorem: v f = v fx 2 +v fy 2 = = 3.89 = 2.0 m/s Use tan θ = (opp / adj ) to find the angle θ. θ = tan 1 ( v y / v x ) = tan 1 ( 1.0 / 1.7 ) = 30 v x = 2.0 m/s [ 30 Down ] Now complete 2e on page 94 22

23 HW/SW: Study examples pp complete #s 20, 22, 27, 29, 31 and 33 pp

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