Chapter 4: 2-D Kinematics

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1 PHY 5 Ch 4. Solution Dr. Hael Shehadeh. Chapter 4: -D Kinematics Answers to Conceptual Questions. The component of velocit is first positive and then negative in a smmetric fashion. As a result, the average component of velocit is zero. The x component of velocit, on the other hand, is alwas v cos. Therefore, the projectile s average velocit has a magnitude of v cos and points in the positive x direction. 4. (a) No. The acceleration is alwas verticall downward, but the fl ball is alwas moving at an angle to the vertical, never straight down. Therefore, its velocit is never vertical and is never parallel to the acceleration. (b) Yes. A projectile at the top of its trajector has a velocit that is horizontal, while at the same time its acceleration is vertical. 6. Just before it lands, this projectile is moving downward with the same speed it had when it was launched. In addition, if it was launched upward at an angle above the x axis, it is moving in a direction that is an angle below the x axis just before it lands. Therefore, its velocit just before landing is v m/s xˆ 4 m/s ˆ.. The tomato lands on the road in front of ou. This follows from the fact that its horizontal speed is the same as ours during the entire time of its fall. 4. Solutions to Problems & Conceptual Exercises The car moves up the 5.5 incline with constant acceleration, changing both its horizontal and vertical displacement simultaneousl. Find the magnitude of the displacement along the incline, and then independentl find the horizontal and vertical components of the displacement. Solution:. (a) Find the magnitude of the displacement along the incline using equation -:. Find the horizontal component of r 3. (b) Find the vertical component r r v t at. m/s s 4 m : x d cos 4 m cos5.5 4 m : d sin 4 m sin5.5 4 m The horizontal and vertical motions can be considered separatel. In this case the are each described b constant acceleration motion, but the vertical acceleration is less than the horizontal. The two accelerations would be equal if the angle of the incline were Two divers run horizontall off the edge of a low cliff. Use a separate analsis of the horizontal and vertical motions of the divers to answer

2 PHY 5 Ch 4. Solution Dr. Hael Shehadeh.. 8. the conceptual question. Solution:. (a) As long as air friction is neglected there is no acceleration of either diver in the horizontal direction. The divers will continue moving horizontall at the same speed with which the left the cliff. However, the time of flight for each diver will be identical because the fall the same vertical distance. Therefore, diver will travel twice as much horizontal distance as diver.. (b) The best explanation (see above) is I. The drop time is the same for both divers. Statement II is true but not relevant. Statement III is false because the total distance covered depends upon the horizontal speed. If air friction is taken into account diver will travel less than twice the horizontal distance as diver. This is because air friction is proportional to speed, so diver, traveling at a higher speed, will experience a larger force. A diver runs horizontall off a diving board and falls down along a parabolic arc, maintaining her horizontal velocit but gaining vertical speed as she falls. Find the vertical speed of the diver after falling 3. m. The horizontal velocit remains constant throughout the dive. Then find the magnitude of the velocit from the horizontal and vertical components. Solution:. Use equation 4-6 to find v :. Use the components vx and v to find the speed: v v g 9.8 m/s m 3. m 58. v v v.85 m/s 58.9 m /s 7.89 m s x Projectile problems are often solved b first considering the vertical motion, which determines the time of flight and the vertical speed, and then considering the horizontal motion. The sparrow falls down along a parabolic arc, maintaining its horizontal velocit but gaining vertical speed as it falls. Find the time it takes the sparrow to travel a horizontal distance of.5 m given that its horizontal velocit remains unchanged at.8 m/s. Then find the distance the sparrow falls during that time interval. Solution:. (a) Find the time to travel x t m.5 m horizontall. v.8.5 m.778 s. Find the vertical drop distance: x s h gt 9.8 m/s.778 s.378 m 3. (b) If the sparrow s initial speed increases, the time interval required for it to travel.5 m horizontall decreases. The distance of fall decreases for a shorter time interval.

3 PHY 5 Ch 4. Solution Dr. Hael Shehadeh. The speed of the sparrow determines v x and gravit determines v. Fling faster will increase v x but not v. 4. The basketball s trajector is depicted at right. Use equation 4-7 to find the x and positions of the basketball as a function of time. Use the right triangle formed b the floor and the basketball s release and landing points to write a ratio that allows us to calculate the time of flight and therefore the initial height. Solution:. Find the position as a function of time: h gt h gt. Find the x position as a function of time: 3. Use the tangent function for the right triangle: 4. Now solve for the flight time t: x v t h gt gt tan x v t v t v tan 4. m/s tan 3. g 9.8 m/s.495 s Find the initial height: h gt 9.8 m/s.495 s. m If the basketball plaer throws the ball from the same height but with a higher initial speed, the 3. angle will decrease. For instance, v 8.4 m/s produces an angle of 6.. Dropping the ball from rest makes the angle 9.. A projectile that is launched at an angle above horizontal follows a parabolic path. The projectile is accelerated onl b gravit, so it maintains its horizontal velocit while its vertical velocit is reduced from a large positive value at launch to zero at the peak of its flight. Therefore the speed of the projectile at the peak of its flight is equal to its horizontal speed at launch. Use this fact to determine the launch angle. Solution: Set vpeak vx and solve for θ: v vpeak v vx v cos cos 6 v If the launch angle were 45 the speed at the peak would be v. 3

4 PHY 5 Ch 4. Solution Dr. Hael Shehadeh. 8. Three projectiles A, B, and C are launched with different initial speeds and angles and follow the indicated paths. Separatel consider the x and motions of each projectile in order to answer the conceptual question. Solution:. (a) Since each projectile achieves the same maximum height, which is determined b the initial vertical velocit, we conclude that all three projectiles have the same initial vertical velocit. That means the larger the horizontal velocit, the larger the total initial velocit. The largest initial speed will therefore correspond with the longest range. The ranking is thus A < B < C.. (b) The flight time is longest for projectiles that have the highest vertical component of the initial velocit. In this case each projectile has the same maximum altitude and therefore the same initial vertical speed. That means the all have the same time of flight and the ranking is thus A = B = C. Projectile C travels the farthest distance in the same amount of time because it has the highest speed. 37. The golf ball travels along a parabolic arc, landing at the same level from which it was launched. The maximum range of a projectile launched from level ground occurs when the launch angle is 45. Use equation 4- to predict the range of the golf ball when launched at 45. The minimum speed of the ball will occur when the ball reaches the peak of its flight. At that point the vertical component of the velocit is zero and the speed equals the horizontal component of the velocit, which remains unchanged throughout the flight. Solution:. (a) Find the range of the ball when it is launched at 45 b using equation 4-:. (b) Find the x component of the ball s velocit, which corresponds to the minimum speed during the flight: 34.4 m/s v R sin sin9 m g 9.8 m/s x v v cos 34.4 m/s cos m/s The maximum range will occur at θ =45 onl in the absence of air resistance. In the presence of the atmosphere ou must launch the ball at a lower angle than that in order to maximize the range of the ball. 4

5 PHY 5 Ch 4. Solution Dr. Hael Shehadeh. 46. The dolphin travels along a parabolic arc as it glides through the air and lands in the water. Because the dolphin is traveling horizontall as it passes through the hoop, we conclude that v at that point and that the dolphin must be at the peak of its flight. Use the formula derived in Example 4-7 to find how high the center of the hoop is above the surface of the water. Solution: Find the maximum height of the dolphin above the water: v sin. m/s sin m g 9.8 m/s max In order to pass through a higher hoop the dolphin must either increase the launch angle or jump with a higher initial speed. 56. The ha bale travels along a parabolic arc, maintaining its horizontal velocit but changing its vertical speed due to the constant downward acceleration of gravit. The initial velocit of the bale is given as v. m/s xˆ 8.85 m/s ˆ. Use the fact that the horizontal component of the bale s velocit never changes throughout the flight in order to find the vertical component of the velocit when the total speed is 5. m/s. Then find the time elapsed between the initial throw and the instant the bale has that vertical speed. For part (b) set the vertical speed equal to the (constant) horizontal speed in magnitude but negative in direction (pointing downward). Find the time elapsed between initial throw and the instant the bale has that new vertical speed. Solution:. (a) Determine the component of the velocit when the total speed is 5. m/s:. Use the positive value of v because the bale is rising when the speed first equals 5. m/s. Find the time elapsed to this point from equation 4-6: 3. (b) If the bale s velocit points 45. below the horizontal then we know the vertical velocit: 4. Find the time elapsed from equation 4-6: v v v x v v v x 5. m/s. m/s v v 4.87 m/s 8.85 m/s t.45 s g 9.8 m/s v v x. m/s v v. m/s 8.85 m/s t. s g 9.8 m/s 5

6 PHY 5 Ch 4. Solution Dr. Hael Shehadeh. 5. (c) If v is pointed straight upward then the initial vertical velocit component will be larger, so it will rise higher and its time in the air will increase. The bale will have a speed of 5. m/s again after.4 s has elapsed. If it were thrown straight upward with the same initial speed (8.9 m/s) it would rise to a height of 4.6 m, as opposed to 3.99 m as in the original case. 7. A ball is thrown at 4. above horizontal and follows a parabolic arc until it returns to the same level from which it was thrown. When the ball returns to the same level from which it was originall thrown, the smmetr of its trajector means that it has the same vertical speed that it had when it was thrown, onl in the downward direction instead of upward. If we know v and the time elapsed, then we can use equation 4- to determine the initial speed of the throw. Solution: Solve equation 4- for the initial speed, letting the final vertical speed v v sin v : v v sin gt v sin v sin gt 9.8 m/s.75 s gt v sin sin 4.. m/s Another wa to approach this problem is to realize that the vertical speeds of the ball thrown straight upward and the ball thrown at 4. must be the same, because the time of flight is determined b the vertical speed, not the horizontal speed. A third wa to approach this problem is to set the vertical speed equal to zero at.375 s after the throw and solve for v as above. Verif for ourself that this ball reaches a maximum height of 9.7 m and lands 44. m downrange. 6