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**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. ...

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