chap3 solutions - CHAPTER THREE SOLUTIONS CHAPTER THREE...

Info icon This preview shows pages 1–4. Sign up to view the full content.

Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER THREE SOLUTIONS CHAPTER THREE SOLUTIONS Chapter Three Readings Brancazio, P, “The Trajectory of a Fly Ball," The Physics Teacher, January 1985, p. 20. Brown, R.A., "Maximizing the Range of a Projectile", The Physics Teacher, 30, 344,1992 Drake, S. and Maclachlan, J., "Galileo's Discovery of the Parabolic Trajectory," Scientific American, March 1975, p. 102. Review (a) The length of the height of the triangle is given by h = 260.8 sin 50° = 199.8 ft. The side DB has length DB = 260.8 cos 50° = 167.6 ft, leaving the side AD = 311.4 - 167.6 = 143.8 ft. AD 143.8 _ — 1 —— - 1 —— _ ° The angle H—tan h atan 199.8 - 35.7 and AC = [(AD)2 + h211/2 = [(143.3)2 + (199.8)2] 1/2 = 246.2 ft. The remaining side of the triangle (from C to A) is thus 246.2 ft in a direction 35.7° west of south. (b) The area of the triangle is given by Area = % bh=% (311.4)(199.8) 1 acre = 4 2 W 2 3.111x10 ft (435 2) 0.714acres. 3.1 Your sketch should be drawu to I N scale, and should look somewhat like 1 5 meters that pictured at the right. The angle from the westward direction, 6, can be measured to be 4.34" north of west, and the distance from your sketch can be converted according to the scale used to be 7.92 m. ........ 3.2 Your sketch when drawn to scale should look somewhat like the one at the right. The distance R and the angle scan be measured to give, upon use of your scale factor, the values of R = 421 ft at about 3° below the horizontal. ”)1 3.8 3.9 3.10 3.11 3.12 3.13 CHAPTER THREE SOLUTIONS with an angle of tan' 12% = 42.0° south of east. The person would have to walk 3.10 sin(25.0“) = 1.31 km north, and 3.10 cos(25.0°) = 2.81 kmeast (3.) Her resultant x (east—west) displacement is -3.00 + 0 + 6.00 = 3.00 blocks, while her resultant y (north-south) displacement is 0 + 4.00 + 0 = 4.00 blocks. Her resultant displacement is then 5.00 blocks at 53.1“ north of east. (b) Her total distance is 3.00 + 4.00 + 6.00 = 13.00 blocks. Let A be the vector corresponding to the 10.0 yd run, B to the 15.0 yd run, and C to the 50.0 yd pass. Also, we choose a coordinate system with the + y direction downfield, and the + x direction toward the sideline to which the player runs. The components of the vectors are then Ax =0 Ay = ‘10.0 Yd 13,( = 15.0 yds By=0 qzo Cy =+_ 500de From these, Rx =21”); =15.0 yds, and Ry =21"); = 40.0 yds, and the Pythagorean theorem gives R=’\/(Rx)2 + (Ry)2 =V(15.0 yds)2 + (40.0 yds)2 =42]de After 3 h moving at 41.0 km/h, the hurricane is 123 km from the island, 61.5 km west and 106.52 km north. It then travels an additional (4.50 - 3.00)(25.0) = 37.5 km north. In total, it has traveled 61.5 km west and 106.52 + 37.5 = 144.02 km north. Therefore, the hurricane is \I(61.5)2 +- (144.02)2 = 156.6 km away from the island. The east and north components of the displacement from Dallas (D) to Chicago (C) are the sums of the east and north components of the displacements from Dallas to Atlanta (A) and from Atlanta to Chicago. In equation form CbCeast = mAeast + @Ceast = 730 C0850 ‘ 560 3.111210 = 727.2 " 200.6 = 526.6 mfles dDCnorth = CbAnorth + dACnorth- = 730 Sin5° + 560 C0821° =- 63.6 + 522.8 = 586.4 miles. By the Pythagorean theorem, d= (dogma?! + .(obmonhfl =788.1 mi Then tan 3:2??? = 1.12 and e = 43.1°. Thus, Chicago is 788 miles at 48.1“ north of east of Dallas. Finding the components of the displacements a, b, and c gives: ax = acos(30.0°) = +152 km, ay = asin(30.0°) = +87.5 km _ bx = bcos(110.0°) = -51.3 km, by = bsin(110.0°) = 141 km ex = ccos(180° = -190 km, Cy = csin(180") = 0 Therefore, the components of the y north 73 3.19 3.20 3.21 3.22 CHAPTER THREE SOLUTIONS «50.0 m = % {-9.80 m/sz) t2, which gives t= 3.19 s. (b) At impact, the horizontal component of velocity is vx =Vox = 18.0 111/3, and the. vertical component is vy = voy + at= 0 + (-9.80 m/s2)(3.19 s) — — 31.3 m/s. The resultant velocity is found from the pythagorean theorem v=\l(31.3 m/s)2 + (18.0 m/s)2 =36.1m/s, at an angle below the horizontal found as tans = 313/180 which yields 9 = 60.1“. We choose our origin at the initial position of the projectile. After 3 s, it is at ground level, y = -H. To find H, we use y= voyt+ % atz. -H= (15 m/s)(sin25°)(3 s) + ‘%(-9.80 m/sz)(3.0 s)2 - -25.1 m, or H= 25 in First, compute the components of the initial velocity. vox= vocos 53.0“ - 12.0 m/s, voy- vosin 53.0” = 16.0 m/s. (a) We can find the time required for the ball to reach the position of the crossbar from X= voxtas: 36.0 m - (12.0 m/s)(t), or t:- 2.99 5. At this time the height of the football above the ground is y- voyn % at2 = (16.0 m/s)(3.005) + —;- (-9.80 m/s2)(3.00 s)2 - 3.90 m. Thus. the ball clears the crossbar by 3.90 m - 3.05 m — 0.85m. (b) The vertical component of the velocity of the ball as it moves over the crossbar is vy = voy + at= 16.0 m/s - (9.80 m/sz)(3.00 s) - -13.4 m/s. The negative sign indicates the ball is moving downward. (a) First, find the speed of the car when it reaches the edge of the cliff frOm V2 = v02 + 2ax= 0 + 2(4.00 m/sz)(50.0 m), or V = 20.0 m/s. Now, consider the projectile phase of the car's motiou. We shall first find the vertical velocity with which the car strikes the water as Vyz = voyz + 22y = ((-20.0 m/s)sin24.0°)2 + 2(-9.80 m/s2){-30.0 m), orvy = -25.6 m/s. The time of flight is found from vy= voy + at as 35.6 m/s = {-20.0 m/s)(sin24.0°) + (-9.80 m/sz)t, which gives t= 1.7 s. The horizontal motion of the car during this time is X=- Voxtz (20.0 m/s)(cos24.0°)(1.78 s) - 32.5 m. (b) The time of flight of the car has already been found in part (a). The components of the initial velocity are vox = 40.0 cos30.0° = 34.6 m/s and V0y= 40.0 sin30.0° = 20.0 m/s The time for the water to reach the building is given by x== vox t,. or 50.0 m- (34.6 m/s) t and t- 1.44 s. Then the height of the water when it reaches the building is given by y= voyr - % th = 20.0(1.44) - 4.9(1.44)2 75 3.27 3.28 3.29 CHAPTER THREE SOLUTIONS Vbs =' V bw + VWS Northward component of Vbs is Vbs cost) = 10.0 m/s (1) Eastward component is V135 sine = 1.50 m/s (2) Dividing (2) by (1) gives tan 9= 0.15 and 9= 8.53”. Then from (1) vb =M= 101 m/s. ’ 3 cos 8.53 ' . . . 300 m 300 m The time to cross the river is t=Vbs c088 = 101) m/s = 30s. The eastward drift = (Vbs sine) t= (1.50 m/s)(30 s) =- 45 m. wa = the velocity of the boat relative to NORTH the water. vws = the velocity of the water relative to the shore, and is directed east. Vbs = the velocity of the boat relative to the shore. Vbs = v bw + M The northward components of this equation (where north is across the stream) are VbslN= wa)N + sz)N Vbs)N = (3.30 mi/h) sin 62.5 + 0 = 2.93 mi/ h. And, the time to cross the stream is 0.505 mi ”2.93 mi/h = 0.172 h. The eastward components of the relative velocity equation (where east is parallel to the current) are: - V1395 = wa)E + sz)E = -(3.30 mi/h) cos 62.5 + 1.25 mi/h = -0.274 mi/h. The distance traveled parallel to the shore = (Vbs)E t= {-0.274 mi/h)(0.l72 h) = 00472 mi = -249 ft = 249 ft upstream. V EAS'I vpa = the velocity of the plane relative to the air = 200 km/h. vag = the velocity of the air relative to the ground = 50.0 km/h (south). vpg = the velocity of the plane relative to W the ground (to be due east). v V pg V pg = V pa + V ag 39 For vpg to have zero northward component, We must have Vpa sine= vag , or sin9= 52%g—km/1111 =025 and 6= 14.5" Thus, the plane should head at 14.S° north of west. Since vpg has zero northward component, vpg = vpacose = (200 km/h)cosl4.5°. This gives, vpg = 194 km/h, the plane's ground speed. The bumpers are initially 100 m.= 0.1 km apart. After time t the bumper of the leading car travels 401‘, while the bumper of the chasing car travels 60:. Since the cars are bumper-to-bumper at time t, we have 0.1 + 40t=60t, yielding t= 0.005 h= 18 s. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern