margin in parentheses. 5. Use the information below to answer parts a-f. Show all your work including vector diagrams, formulas, etc. A soccer ball is kicked at an angle of 32.0 above the horizontal with an initial velocity of 25.0 m/s. a) Determine the horizontal component of the initial velocity. (1) V=25 Voy-295132) (1) * Vox = 25 cos (32) 21.20 M/S 21.2ml5 Von: 25 cc (32) b) Determine the vertical component of the initial velocity. Voy = 25 sin (32) ~ 13.25 m/s 13.3MLS c) What is the ball's horizontal velocity at 2.00 s? (1) There is no horizontal acceleration so Vox is Constant as theres nothing in the horizontal direction to change its velocity So the Velocity is 21.20m (s. d) What is the ball's vertical velocity at 2.00 s? 4. A ball rolls off a roof with a horizontal velocity of 1.75 m/s. The ball lands 3.50 m from the edge of the building. How tall is the building? Show all work. (3) inicial velocity= horizontal, so is Voy = 0 Velochy only in 2. as 2/3 On = Voxt time z = On = 3.5 Von coordinate? 1.75 Ay = 1/2g+3 = 4.9 (3575) 2 The building is 19.6m tall. 14.6m = After you have completed all questions in Lesson 3. please place your assignment in a safe 3 of 6 hysics 20: Mrs. Tyler Assignment 4 Unit A Module 2, Lesson 4: Projectiles in Two Dimensions Lesson 2 of this assignment is worth 15 marks. The value of each question is noted in the left margin in parentheses. 5. Use the information below to answer parts a-f. Show all your work including vector diagrams, formulas, etc. A soccer ball is kicked at an angle of 32.0 above the horizontal with an initial velocity of 25.0 m/s. a) Determine the horizontal component of the initial velocity. Von = 25 cos (32) 21.20 M/S (1) V = 25 Voy=295in(32) 0.75 Von: 25 cc (32) (1) 0.75 sig digs b) Determine the vertical component of the initial velocity. Voy = 25 sin (32) ~ 13.25 m/s sig dig Coordinate? c) What is the ball's horizontal velocity at 2.00 s? There is no horizonal acceleration so Von is in the horizontal direction to change its velocity So the Velocity is 21.20m (s. d) What is the ball's vertical velocity at 2.00 s? (1) Vfy = Voy - gt Vty = assin (22) - (9.8) (2)x-6.35m/s ADLC All Rights Reserved (2019) constant as theres nothing 2 3 I am not sure what your answer is. Make sure you are using formulas that you learned about in this course. 6. Use the following information to answer parts a-b. Assume air resistance is negligible and show all work. Tiger Woods' golf balls leave his club with an average speed of around 76.0 m/s. He hits a ball with an angle of 15.0 relative to the horizontal. a) How long is the golf ball in the air? 1/3 (3) JO=76MIS Dy=0 At=vfy-viy/a ave y See Lesson 4, P5 "Projectile Motion Example" 0 = Joy =765in (15) Joyz = von = 7014 (15) voyt-1/28 4/2g+2 2 By = voy=-1/2gt 1- 2003 2(765(15)) = ~2-005 9.8 ADLC All Rights Reserved (2019) 5 Physics 20: Mrs. Tyler Assignment 4 2/2 (2) b) What is the horizontal displacement of the golf ball? Dn =Vont = (7bccs (15)) (2) 146.82m Correct process 6. Use the following information to answer parts a-b. Assume air resistance is negligible and show all work. Tiger Woods' golf balls leave his club with an average speed of around 76.0 m/s. He hits a ball with an angle of 15.0 relative to the horizontal. a) How long is the golf ball in the air? (3) Physics 20: Mrs. Tyler Assignment 4 * b) What is the horizontal displacement of the golf ball? (2) ADLC All Rights Reserved (2019) 4 of 6 5 4. (3) A ball rolls off a roof with a horizontal velocity of 1.75 m/s. The ball lands 3.50 m from the edge of the building. How tall is the building? Show all work. time initial velocity - horizonted, so Voy = 0 Velochy is only 12. as on = Voxt z = by = 3.5 Von 1.75 Ay = 4/2gt = 4.9 (3.5 175) 2 The building i 1963 19.6m 7all. After you have completed all questions in Lesson 3, please place your assignment in a safe place until you have finished Lesson 4.