Physics 100A Homework 3 Chapter A sailboat runs before the wind with a constant speed of 4.5 m/s in a direction 32 north of west.

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1 Phsics A Homework 3 Chapter 4 4. A sailboat runs before the wind with a constant speed of 4.5 m/s in a direction 3 north of west.. Picture the Problem: The ector inoled in this problem is depicted at right. Strateg: Separate r into - and -components. Let north be along the -ais, west along the ais. Find the components of the elocit in each direction and use them to find the distances traeled. Solution:. (a) Find and :. Find the westward distance traeled: 3. (b) Find the northward distance traeled: = ( 4. m/s) cos3 = 3.56 m/s = ( 4. m/s) sin 3 =.3 m/s = t = ( 3.56 m/s)( 5 min 6 s/min) = 53 m = 5.3 km (.3 m/s)( 5 min 6 s/min ) = t = = 33 m = 3.3 km Insight: The northward and westward motions can be considered separatel. In this case the are each described b constant elocit motion. W 3 r N 4.7 Two canoeists start paddling at the same time and head toward a small island in a lake, as shown in the figure. Canoeist paddles with a speed of.65 km/s at an angle of 45 north of east. Canoeist starts on the opposite shore of the lake, a distance of.5 km due east of canoeist. a a Isosceles triangle Strateg: Canoeist s 45 path determines an isosceles right triangle whose legs measure. km. So canoeist s path determines a right triangle whose legs measure. km north and.5 km west. Use the right triangle to find the angle θ. Find the distance traeled b each canoeist and set the times of trael equal to each other to determine the appropriate speed of canoeist : Solution:. (a) Find the angle θ from the right triangle of canoeist.5 km θ = tan = 7. km. (b) Set the trael times equal to each other: Δr Δr t = t = 3. Use the resulting ratio to find the appropriate speed of canoeist : d ( ) ( ) d.5km +.km m = =.35. m/s = s (.km) + (.km) Insight: There are other was to sole this problem. For instance, because the motions are independent, we could set the time it takes canoeist to trael. km horizontall equal to the time it takes canoeist to trael.5 km horizontall. 4 45

2 Introduction to projectile motion. r Consider a particle with initial elocit that has magnitude of. m/s and is directed 6. aboe the negatie ais. 6 Part A What is the component of r? = cos θ = cos(6) = 6 m/s Part B What is the component of r? = sin θ = sin(6) =.4 m/s Part C The ertical component ehibits motion with constant nonzero acceleration, whereas the horizontal component ehibits constant-elocit motion. Part D How long does it take for the balls to reach the ground? Use m/s for the magnitude of the acceleration due to grait. = + o =5+- () t 5t = 5 t = s Part E Imagine that the ball on the left is gien a nonzero initial elocit in the horizontal direction, while the ball on the right continues to fall with zero initial elocit. What horizontal speed must the ball on the left start with so that it hits the ground at the same position as the ball on the right? From the applet we find that the balls are initiall separated a distance of 3. m. The ball on the left must be able to coer this distance as it falls, so that both balls can reach the same point at the bottom. 3 = t = = = 3. m/s t Horizontal cannonball on a clift. A cannonball is fired horizontall from the top of a cliff. The cannon is at height H = 8. m aboe ground leel, and the ball is fired with initial horizontal speed. Assume acceleration due to grait to be g = 9.8 m/s.

3 Part A Assume that the cannon is fired at time t= and that the cannonball hits the ground at time t g. What is the position of the cannonball at the time /? Starts with zero elocit in the -direction and with height H. At the ground: = + o H (8) = H + g t g = = = 4. 4 s g 9.8 At t g / =4.4/ =. s = (.) = 6 m Part B Gien that the projectile lands at a distance D = 3 m from the cliff, as shown in the figure, find the initial speed of the projectile. D 3 = t D = tg = = = 3. m/s t 4.4 g Part C What is the position of the cannonball when it is at distance D/ from the hill? If ou need to, ou can use the trajector equation for this projectile, which gies in terms of directl: g 9.8(65) = H = 8 = 6 m (3.) 4.8 Two diers run horizontall off the edge of a low cliff. Dier runs with twice the speed of dier. 8. Picture the Problem: Two diers run horizontall off the edge of a low cliff. t g Strateg: Use a separate analsis of the horizontal and ertical motions of the diers to answer the conceptual question. Solution:. (a) As long as air friction is neglected there is no acceleration of either dier in the horizontal direction. The diers will continue moing horizontall at the same speed with which the left the cliff. Howeer, the time of flight for each dier will be identical because the fall the same ertical distance. Therefore, dier will trael twice as much horizontal distance as dier.. (b) The best eplanation (see aboe) is I. The drop time is the same for both diers. Statement II is true but not releant. Statement III is false because the total distance coered depends upon the horizontal speed. Insight: If air friction is taken into account dier will trael less than twice the horizontal distance as dier. This is because air friction is proportional to speed, so dier, traeling at a higher speed, will eperience a larger force.

4 4. A dier runs horizontall off the end of a diing board with an initial speed of.5 m/s. If the diing board is 3.5 m aboe the water, what is the dier's speed just before she enters the water?. Picture the Problem: A dier runs horizontall off a diing board and falls down along a parabolic arc, maintaining her horizontal elocit but gaining ertical speed as she falls. Strateg: Find the ertical speed of the dier after falling 3. m. The horizontal elocit remains constant throughout the die. Then find the magnitude of the elocit from the horizontal and ertical components. Solution:. Use equation 4-6 to find :. Use the components and g ( )( ) to find the speed: ( ) = Δ = 9.8 m/s m 3. m = 58.9 m /s = + =.85 m/s m /s = 7.89 m s Insight: Projectile problems are often soled b first considering the ertical motion, which determines the time of flight and the ertical speed, and then considering the horizontal motion. Arrow hits the apple θ = 45 o An arrow is shot at an angle of aboe the horizontal. The arrow hits a tree a horizontal distance D= m awa, at the same height aboe the ground as it was shot. Part A Find, the time that the arrow spends in the air. When the arrow hits the ground at the same height it was shot the magnitude of its final elocit is the same as that of its initial elocit. The component of the elocit points opposite. The -equation = t D = cos(45) t = D cos(45) t The elocit equation = = = o o o = sin(45) We can now equate both equations (recall cos(45)=sin(45), so these cancel) sin(45) = D cos(45) t D () t = t = = 6. 7 s g 9.8 Part B Suppose someone drops an apple from a ertical distance of 6. meters, directl aboe the point where the arrow hits the tree. How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree? Since the arrow hits the tree at the ground, then we require that the apple hit the ground at the same moment.

5 The time for the apple to hit the ground from a height of 6. m haing started from rest is: = + o 6 9.8t (6) 9.8 = + t = =. s So if the arrow is in flight 6.7 s, the apple must be dropped 6.7-.=5.6s after the arrow is shot. Projectile Motion Conceptual Part A Based on the equations of motion the projectile follows a parabolic path, B Part B The acceleration is alwas pointing downwards with a constant alue of 9.8 m/s Projectile Motion Ranking Task Part A The case achieing the maimum height Part B The case taking the longest time to hit the ground In both cases the point is to find which case has the largest initial elocit in the direction - = 5sin(6) = 3 m/s - = sin(6) = 8.7 m/s 3- = sin(9) = m/s 4- = 5sin() = m/s 5- = 5sin(3) = 7.5 m/s 6- = 5sin(45) =.6 m/s The order from largest to smallest:, 6, 3,, 5, 4

6 4.8 Three projectiles (A, B, and C) are launched with different initial speeds so that the reach the same maimum height, as shown in the figure. List the projectiles in order of increasing initial speed. 8. Picture the Problem: Three projectiles A, B, and C are launched with different initial speeds and angles and follow the indicated paths. Strateg: Separatel consider the and motions of each projectile in order to answer the conceptual question. Solution:. (a) Since each projectile achiees the same maimum height, which is determined b the initial ertical elocit, we conclude that all three projectiles hae the same initial ertical elocit. That means the larger the horizontal elocit, the larger the total initial elocit. 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 hae the highest ertical component of the initial elocit. In this case each projectile has the same maimum altitude and therefore the same initial ertical speed. That means the all hae the same time of flight and the ranking is thus A = B = C. Insight: Projectile C traels the farthest distance in the same amount of time because it has the highest speed. 4.9 A second baseman tosses the ball to the first baseman, who catches it at the same leel from which it was thrown. The throw is made with an initial speed of 8. m/s at an angle of 35. aboe the horizontal 9. Picture the Problem: The ball traels along a parabolic arc, maintaining its horizontal elocit but changing its ertical speed due to the constant downwards acceleration of grait. Strateg: The gien angle of the throw allows us to calculate the horizontal component of the initial elocit b using the cosine function. The ertical component of the elocit can be found b using the sine function. The time it takes the acceleration of grait to slow down the ertical speed, bring it to zero, and speed it up again to its initial alue is the same as the time the ball is in the air. Solution:. (a) Find the component of the initial elocit: θ ( ) = cos = 8. m/s cos37.5 = 4.3 m/s. (b) Find the component of the initial elocit: θ ( ) 3. Let and use equation 4-6 to find the time of flight: = sin = 8. m/s sin 37.5 =. m/s =. (. m/s) t = = = g 9.8 m/s.4 s Insight: The flight of the ball is perfectl smmetric the angle of the motion is 37.5 below horizontal at the instant it is caught, and the ball spends the same amount of time going upward to the peak of its flight as it does coming downward from the peak. This is onl true if the ball is caught at the same leel from which it was thrown.

7 4.3 A cork shoots out of a champagne bottle at an angle of 34. aboe the horizontal. If the cork traels a horizontal distance of.5 m in.35 s, what was its initial speed? 3. Picture the Problem: The cork traels along a parabolic arc, maintaining its horizontal elocit but changing its ertical speed due to the constant downward acceleration of grait. Strateg: Find the horizontal component of the initial elocit b diiding the horizontal distance traeled b the time of flight. Then use the cosine function to find the initial speed of the cork..3 m Solution:. Find the horizontal speed of the cork: = = =.4 m/s = t.5 s. Use the cosine function to find the initial speed: cosθ.4 m/s cos35. = = =.7 m/s Insight: Because grait acts onl in the ertical direction, the horizontal component of the cork s elocit remains unchanged throughout the flight The "hang time" of a punt is measured to be 4.5 s. If the ball was kicked at an angle of 6. aboe the horizontal and was caught at the same leel from which it was kicked, what was its initial speed? 39. Picture the Problem: The football traels along a parabolic arc, landing at the same leel from which it was launched. Strateg: The time of flight of a projectile that lands at the same leel it is launched is determined b the time it takes the acceleration of grait to slow down the ertical component of the initial elocit to zero and then speed it up again back to its original alue. Thus upon landing the speed of the ball is sin = = θ. Use these facts to determine the time of flight and then sole for. Solution:. Find the time of flight: sinθ sinθ sinθ t = = = g g g. Sole for : ( 9.8 m/s )( 4.5 s) = = = 4.8 m/s sinθ sin63. Insight: The flight of the football would not be perfectl smmetric if air resistance were present or if it were caught at a different leel from which it was kicked. 4.4 In a friendl game of handball, ou hit the ball essentiall at ground leel and send it toward the wall with a speed of 4 m/s at an angle of 38 aboe the horizontal. Part A How long does it take for the ball to reach the wall if it is 3.4 m awa? 4. Picture the Problem: The path of the ball is depicted at right. Strateg: Use the horizontal component of the ball s elocit together with the horizontal distance d to find the time elapsed between the hit and its collision with the wall. Then use the time to determine the ertical position h of the ball when it collides with the wall. Solution:. (a) Use equation 4- d t = = 3.8 m 8 m/s cos3 to find the time: cosθ ( ) =.5 s. (b) Use equation 4- to find h: h = ( sinθ ) t ( ) ( ) ( )( ) = 8 m/s sin 3.5 s 9.8 m/s.5 s =. m Insight: In man cases the ertical motion determines the time of flight, but in this case it is the horizontal distance between the point where the ball is struck and the wall that limits the time of flight. r θ d h

8 4.43 On a hot summer da, a oung girl swings on a rope aboe the local swimming hole. When she lets go of the rope her initial elocit is. m/s at an angle of 35. aboe the horizontal. If she is in flight for.66 s, how high aboe the water was she when she let go of the rope? 43. Picture the Problem: The trajector of the girl is depicted at right. Strateg: Use equation 4- and the gien time of flight, initial speed, and launch angle to determine the initial height of the girl at the release point. Solution: Use equation 4- to find the initial height of the girl at the release point. If we let the release height correspond to =, then the landing height is: ( sinθ ) (.5 m/s)( sin 35. )(.66 s) ( 9.8 m/s )(.66 s) = t = =.7 m In other words, she was.7 m aboe the water when she let go of the rope. Insight: The girl s speed upon impact with the water is 5. m/s (check for ourself) or mi/h. A fun plunge!