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'I in” 3.8. 10. mmwmw'mmdemmammmmmnmgmmmuqe Thus.
theymachthegmmmwiththesameverﬁcalmed. However,thebaudnownhonnunlaﬂyhadanmhal
honzonmloomponemofvelocitywhichismsimainedthrmghmnﬂiemotion Thus,tbebslltiuown
honmnmﬂymwiththegreezerweed. 3.3 [3) Drawing these to scale and maintaining their respective directions yields a resultant of 5.2 m at an angle of 60“ above the x axis. (b) Maintain the direction of A, but reVerse the direction of B by 180'.
The resultant is 3.0 m at an angle of 30° below the x axis. (c) Maintain the direction of B, but reverse the direction of A. The
"resultantis 3.0 matanange 0f 150°withrespectto'the +xaxis. (d) Maintain the direction of A, revase the direction of B, and multiply
its magnitude by two. The resultant is 5.2 m at an angle of 60° below
the + X axis. The person would have .to walls _ _ _ __
3.10 sin{25.0°) = 1.31 km north, and 3.10 cos‘(25.0°)  2.81kmeast. '3 16 Vox = 100.8 m1/li= 45.06 m/s and the distance traveled in= 60.0 ft = 18.29 m.
. The time to reach homeplate is given by x voxt , which yields
18.29 mi'(45.06 m/S)t_ , or t=0.406 s. _
InthisﬁmeintervaLtheballhasfallenadistance of 1 . =
y .= v9yt+ % atZ  0 + 2 {—9.80 m/s2)(0.406 s)2 = {1.307 m 2.65 ft. u _— 3.13 (a) vex18.01%, Voy=0.Weﬁndthetimeoffallas yavoynéatl, or 50.0 m % (9.30 m/sz) :2, which gives ca 3.19 s. (h) At impact, the horizontal component 'of "velocity is vx vox  18.0
m/s, and the vertical component '15 ‘
v = voy + at= 0 + {9.80 .m/s2)(3.19 s) a  31.3 m/s. 0:
Theyresultant veloci is found from the .pythagOrean  eorem a 31.3 m/s)2 + (18.0 m/s)2 =36.'1m/s,
at :11 angle below the horizontal found as
tan9= 313/180 whichyields 9601. 3.23 (a) The projectile is moving horizontally at the highest point at the
trajectory. The y—component of velocity is zero but the X~component is
voxvocos 00 60cios30° .. 52.0 m/‘s
(b) The x—component of the distance is vo(cos 90) t 60(cos 30°) 4 ... 208 in.
The y—component of the distance is vo(sin 90) r ~§ gtl  60(sln 30°) 4  4.9(4)2 = 41.6 m.
The distance is given by the Pythagorean theorem to be 212 m.
3.24 The velodty of the plane relative to
the mound is the vector sum of the “1011’“ velocity of the plane relative to the
air and the velocity of the air relative u“ totheground, or VDE‘V_pa+Vag. I swig10011911
The components of this velocity are __ _ _ _ _. _ ° Vp)east == "300 + 100 cos30.0“ = 387 mph v». 300 mm In
and vﬂmmh = 0 + 100 8in30.0° = 50 mph Thus, the ma ‘tude and direction are given by
Vp = qupeast .+ Vzpnoﬁjh =390 mph and tan 9m=0.129 and a= 7.4“ ' ‘ Vpﬂst _
The plane moves at 390 mph at 7.4“ north of east relative to the ground. ...
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 Fall '01
 Shapiro
 Physics

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