Vectors and Projectiles Projectile Motion

Vectors and Projectiles Name: Projectile Motion Read from Lesson 2 of the Vectors and Motion in Two-Dimensions chapter at The Physics Classroom: MOP Connection: Vectors and Projectiles: sublevels 7 - 10 1. A baseball is dropped off a cliff and it accelerates to the ground at a rate of -9.8 m/s2, down. Meanwhile a cannonball is launched horizontally from a cannon with a horizontal speed of 20 m/s. 2. A scale is shown along the sides of the graphic at the right. Use the scale to locate the position of the baseball and the cannonball. Trace a line to indicate the trajectory of the cannonball. Baseball Cannonball t (s ) y (m) t (s) 0 1 0.0 4.9 2 3 4 5 0 1 19. 6 44. 1 78. 4 122 .5 3. Which of these two balls strike the ground first? They hit the ground at the same time. 4. Compare the two diagrams - the vertical free-fall motion on the left and the twodimensional free-fall motion on the right. Describe the effect on an object's horizontal motion upon the object's vertical motion. x (m) 0.0 20. 2 3 4 5 0 40. 0 60. 0 80. 0 100 .0 The horizontal motion has no effect upon the vertical motion; perpendicular components of motion are independent of each other. Page 1 y (m) 0.0 4.9 19.6 44.1 78.4 122.5 Vectors and Projectiles 5. The diagram below shows the trajectory of a horizontally launched projectile. Positions of the projectile at 1-second intervals are shown. Demonstrate your understanding of the components of the displacement vector by determining the horizontal displacement (x) and the vertical displacement (y) after the fifth second. 6. A ball is launched horizontally from the top of a cliff with an initial velocity of 20 m/s. The trajectory of the ball is shown below. Express your understanding by filling in the blanks. x= 250 m. right y= 125 m, down 20 m/s 9.8 m/s 20 m/s 19.6 m/s 20 m/s 29.4 m/s 20 m/s 39.2 m/s 7. If the ball in the diagram above strikes the ground after four seconds, then (a) how high was the cliff and (b) how far from the base of the cliff will the ball land? PSYW (a) The distance fallen is given by the equation dy = voy•t + ½•ay•t2 where voy = 0 m/s, ay = -9.8 m/s2, and t = 4.0 s. By substitution, dy = -78.4 m. The cliff is 78.4 m high. (b) The horizontal displacement is given by the equation dx= vox•t where vox = 20 m/s and t = 4.0 s. By substitution, dx = 80 m. The ball lands 80 m from the base of the cliff. 8. If the ball's initial speed in question #6 was 16 m/s, then how far from the cliff will the ball land? Since an alteration in the horizontal velocity does not affect the time to fall vertically, the ball will still be in the air for 4.0 seconds. Using the dx= vox•t, and substituting 16 Page 2 Vectors and Projectiles Name: m/s for the new vox, it can be determined that the ball lands 64 m from the base of the cliff. Page 3 Vectors and Projectiles 9. Use the diagram below to construct a free-body diagram for a vertically launched projectile as it rises towards its peak, at its peak, and as it is falls from its peak. Fgrav Fgrav Fgrav 10. Use the diagram below to construct a free-body diagram for a projectile launched at an angle as it rises towards its peak, at its peak, and as it is falls from its peak. Fgrav Fgrav Fgrav 11. A projectile is launched with a speed of 31.1 m/s at an angle of 71.2 degrees above the horizontal. The horizontal and vertical components of the initial velocity are shown in the first row of the data table. Fill in the table indicating the value of the horizontal and vertical components of velocity for the projectile during the course of its motion. 0.0 10. 0 10. 0 10. 0 10. 0 10. 0 10. 0 -9.8 19. 6 9.8 0.0 -9.8 19. 629. 4 Page 4 Vectors and Projectiles Name: The Equations: Kinematic equations used for 1-dimensional motion can be used for projectile motion as well. The key to their use is to remember that perpendicular components of motion are independent of each other. As such, the equations for one dimension must be applied to either the horizontal motion of a projectile or the vertical motion of a projectile. When using the equations to analyze projectile motion, one assumes negligible air resistance and an acceleration of gravity of 9.8 m/s2, down(-). Thus, ax = 0 m/s/s and ay = -9.8 m/s/s. 1-Dim. x-comp. y-comp. vf = vo + a•t vfx = vox + ax•t vfy = voy + ay•t d = vo•t + •a•t2 vf2 = vo2 + 2•a•d dx = vox•t + •ax•t2 vfx2 = vox2 + 2•ax•dx dy = voy•t + •ay•t2 vfy2 = voy 2 + 2•ay•dy vo d= vf 2 vox dx = •t v fx 2 voy dy = v fy 2 12. A ball is projected horizontally from the top of a 92.0-meter high cliff with an initial speed of 19.8 m/s. Determine: (a) the horizontal displacement, and (b) the final speed the instant prior to hitting the ground. (a) The solution begins by using the equation dy = voy•t + ½•ay•t2 where voy = 0 m/s, ay = -9.8 m/s2, and dy = 92.0 m in order to find the time (t). By substitution, t is found to be 4.333 ... s. Once found, the equation dx = vox•t can be used to determine the horizontal displacement (dx). Substituting in 19.8 m/s for vox yields a dx value of 85.8 m. (b) The final speed is the resultant of two independent velocities - the vfx and the vfy - at the moment the projectile reaches the ground - at 4.333 ... seconds. The vfx value is the same as the vox value - 19.8 m/s - since horizontal velocity does not change. The vfy value is 42.464 ... m/s as found from the equation vfy = voy + ay•t. Once the components are found, they can be added together as vectors using Pythagorean theorem to determine a final speed value of 46.9 m/s. 13. Determine the launch speed of a horizontally launched projectile that lands 26.3 meters from the base of a 19.3-meter high cliff. The solution begins by using the equation dy = voy•t + ½•ay•t2 where voy = 0 m/s, ay = -9.8 m/s2, and dy = 19.3 m in order to find the time (t). By substitution, t is found to be 1.984 ... s. Once found, the equation dx = vox•t can be used to determine the launch speed (vox). Substituting in 26.3 m for dx yields a vox value of 13.3 m. 14. A soccer ball is kicked horizontally at 15.8 m/s off the top of a field house and lands 33.9 metes from the base of the field house. Determine the height of the field house. The solution begins by using the equation dx = vox• t + ½•ax•t2 where vox = 15.8 m/s, ax = 0.0 m/s2, and dy = 33.9 m in order to find the time (t). By substitution, t is found to be 2.145 ... s. Once found, the equation dy = voy•t + ½•ay•t2 can be used to determine the vertical displacement (dy). Substituting values of voy = 0 m/s, ay = -9.8 m/s2, and t = 2.145 ... s, yields a dy value of 22.556 ... m. The cliff is 22.6 m high. Page 5 •t •t Vectors and Projectiles 15. A ball is projected at an angle with an initial horizontal velocity of 8.0 m/s and an initial vertical velocity of 29.4 m/s. The trajectory diagram shows the position of the ball after each consecutive second. Express your understanding of projectiles by filling in the blanks. 9.8 m/s 8 m/s 8 m/s 8 m/s 19.6 m/s 8 m/s 9.8 m/s 8 m/s 19.6 m/s 8 m/s 29.4 m/s 16. Determine … (a) … the displacement of the ball, (b) … the height above the ground at its peak, and (c) … the final speed of the ball upon hitting the ground. (a) The ball is in the air for 6.0 seconds (count the images). It is traveling horizontally at 8 m/s for this entire time. Using the equation dx = vox•t on can determine a horizontal displacement of 48 m. (b) The ball is at its at 3.0 seconds. It's distance above the launch location is given by the equation dy = voy•t + ½•ay•t2 where voy = 29.4 m/s, ay = -9.8 m/s2, and t = 3.0 s. By careful substitution and solving, the vertical displacement at 3.0 seconds (i.e., peak height) is 44 m (44.1 m). (c) The final speed is the resultant of two independent velocities - the vfx and the vfy - at the moment the projectile reaches the ground - at 6.0 seconds. The vfx value is the same as the vox value - 8.0 m/s - since horizontal velocity does not change. The vfy value is -29.4 ... m/s as shown in the diagram. The Pythagorean theorem can be used to add these two perpendicular components. The resultant velocity at 6.0 seconds is SQRT( (8.0 m/s)2 + (-29.4 m/s)2 ) = 30.5 m/s. 17. Suppose that the horizontal component of the initial velocity had been 13.0 m/s and the vertical velocity had been unchanged (in questions #15 and #16). Determine the … (a) … time of flight, (b) … the displacement of the ball, (c) … the height above the ground, and (d) … the speed upon hitting the ground. (a) The time of flight would still be 6.0 seconds since an alteration in the horizontal motion does not affect the time to rise and fall vertically. (b) The horizontal displacement (dx) is affected by this new horizontal velocity. The new value for dx is given by the equation dx = vox•t = (13 m/s)•(6.0 s) = 78 m. (c) The height above the ground would still be 44 m (44.1 m) since the vertical motion is not affected by the alteration in the horizontal motion. Page 6 Vectors and Projectiles Name: (d) Once more, the final speed is the resultant of two independent velocities - the vfx and the vfy - at the moment the projectile reaches the ground. This value is found by adding the 13.0 m/s and 29.4 m/s using Pythagorean theorem. So vf = SQRT( (13.0 m/s)2 + (-29.4 m/s)2 ) = 32.1 m/s. Page 7 Vectors and Projectiles 18. A physics student is driving his pick-up truck down Lake Avenue. The pick-up is equipped with a projectile launcher that imparts a vertical velocity to a water-filled rubber projectile. While traveling 20.0 m/s in an eastward direction, the projectile is launched vertically with a velocity of 58.8 m/s. Fill in the table at the right. showing the horizontal and vertical displacement of the projectile every second for the first 12 seconds. dx = vox•t + •ax•t2 dy = voy•t + •ay•t2 v t (s) dx (m) 0 0.0 dy (m) 0.0 1 20.0 53.9 2 40.0 98.0 3 60.0 132.3 4 80.0 156.8 5 100.0 171.5 6 120.0 176.4 7 140.0 171.5 8 160.0 156.8 9 180.0 132.3 10 200.0 98.0 11 220.0 53.9 12 240.0 0.0 19. On the diagram below, place a large dot on the location of the projectile during each second of its trajectory. Draw a smooth curve through the dots to indicate the trajectory. 20. Will the projectile land in the truck, behind the truck or in front of the truck? (Assume no air resistance.) In the truck! Explain your answer. The projectile maintains a constant horizontal velocity, just like the truck. Its vertical motion (up and down) does not affect the horizontal motion; and so the projectile is at all points in its trajectory directly above the launcher. Page 8 Vectors and Projectiles Name: 23. Use trigonometric functions to resolve the following velocity vectors into horizontal and vertical components. Then utilize kinematic equations to calculate the other motion parameters. Be careful with the equations; be guided by the principle that "perpendicular components of motion are independent of each other." A long jumper leaps with an initial velocity of 9.5 m/s at an angle of 40° to the horizontal. Megan Progress, GBS golf standout, hits a nine-iron with a velocity of 25 m/s at an angle of 60° to the horizontal. A place kicker launches a kickoff at an angle of 30° to the horizontal and a velocity of 30 m/s. vox = 7.277 … m/s vox = 12.5 m/s vox = 25.98 … m/s voy = 6.106 … m/s voy = 21.65 … m/s voy = 15.00 … m/s tup = 0.6231 … s tup = 2.209 … s tup = 1.530 … s ttotal = 1.246 … s ttotal = 4.418 … s ttotal = 3.061 … s dx = 9.069 … m dx = 55.23 … m dx = 79.53 … m dy @ peak = 1.902 … m dy @ peak =23.91 … m dy @ peak =11.47 … m 24. Generalize the calculations performed in question #23 above by writing the equations used to calculate each of the quantities requested in the problem. vox = vo•cosine( ) voy = vo•sine( ) tup = voy / 9.8 ttotal = 2•tup = voy / 4.9 dx = vox •ttotal dy @ peak = voy• tup / 2 25. Determine the range of a ball launched with a speed of 40.0 m/s at angles of (a) 40.0 degrees, (b) 45.0 degrees, and (c) 50.0 degrees from ground level. PSYW and label your answers. The simplified equations from question #24 can be used to calculate the following quantities. The dx values (fifth row of the table) represent the answers to this question. = 40° vox = 30.641 … m/s voy = 25.711 … m/s tup = = 45° = 50° vox = m/s 28.284 … vox = m/s 25.711 … 2.623 … s voy = m/s 28.284 … voy = m/s 30.641 … ttotal = 5.247 … s tup = 2.886 … s tup = 3.126 … s dx = 160.78 … m ttotal = 5.772 … s ttotal = 6.253 … s dx = 163.26 … m dx = 160.78 … m dy @ peak = 33.728 … m dy @ peak = 40.816 … m dy @ peak = 47.904 … m Page 9 Vectors and Projectiles 26. For the three initial launch angles in question #25, determine the peak heights. PSYW and label your answers. Answers to this question can be found in the sixth row of the table above. Page 10