sol to assignment3

# sol to assignment3 - ii'I in” 3.8 10...

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Unformatted text preview: ii 'I in” 3.8. 10. mmwmw'mmdemmammmm-mnmgmmmuqe 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°-) - 2.81-kmeast. '3 16 V-ox = 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 m-i'(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) vex-18.01%, Voy=0.Weﬁndthetimeoffallas ya-voynéatl, 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 9-601. 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 vox-vocos 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 swig-10011911 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 9-m=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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