Question

A helicopter starts from rest at point A and travels along the straight-line path with a constant acceleration a. If the speed v = 33 m/s when the altitude of the helicopter is h = 44 m, determine the values of r, i, o, and as measured by the tracking device at O. At this instant, 0 = 38°, and the distance d = 159 m. Neglect the small height of the tracking device above the ground. ih А d

Answer 1

Given Data V = 33 m/s h = 44 m 0 = 38° d = 159 m To find i opel 20.00 Solution Fiest of all we need to draw FBD of given fig- 1:33 mises B 111. € 371.47m. 1h 44m. 50738 (« = 23. 1979 d=15am. from this frig we no need to calculate AB 4 €! asino be sons Sin(38) 44 w هه sin(389 &=71.47m PAGE-1

Now, I (AB) = ? l(AB) - Toeſt64)'-2X0A XOBC850 By cosine Rule. .: I(78) = €+d2-2dx€ Xcos 38 E55147)+(159)?- 2X159x#147 XC0838 ILAB 111.71m Now, calculate a) using sine Elle, AB sino OB Sina 111.71 71,47 kroon sin 38 sina. x = 23.197° Now, Helicoptoe travelling along straight line path thats why accelenation of helicopter will be - (using Hoear Kinematte eg") 62 u² =zas. 2 33 - 0²=20x111.7 m 29 = 9.75 a = 4.874 m/s PAGE 2

Now, Draw velocity of acceleration vectors. vo Ve aur ao 0-38 a Qt38 213.17 Now, Consider velocity component and radial velocity will be - Ve = vsin (go-0-x) Ve = 33 sin (90-438 - 23. 197 23.1970 VE - 15.9 m/sec ve means - į Ans Now on Velo velocity vo=8 - 3 - 15-9 m)see Ě = -15.9 m/see perpendicul endicular Vo = VC09 (90-0-0) = 33C05 (90 - 38-23:197) Vo - 28.91 mise but vo= { 28.91 H.47 x 0 0,4045 Ead see. Austin o

e Now, considen acceleradion component :: Radial accelenation (de) qera sin (90-0-0) =4874 Sin (90-38-23,197) de = 2.3483 m/ see? bet qe=é + E02 2,3483 = 3 + 71:47X60,4045) = -9,35 Pols? And in Now, fordiserepe accelenirion (am) perpendicular (90-0-6) F41874. Cos (90-38-23:197) (qa = 412709 m/sec) Beut, ao Eo ta é = 4,2709–71,47%+2X415.9)x(0:4045) 4.270g = 71.47 © - 12.863) 71147 6 = 17.134 09 = 0,2397 ead/sec? Ansive AIPALES To acos

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