Question

Ionsider the ski jump track shown in Figure 2 below. The radius of curvature of the track is r= 60 [m] and the ski jumper's mass is 65 [kg]. The ski jumper's angular displacement, 6 is defined from the horizontal as shown. The ski jumper goes along the ski jump track for 6 = 120° before launching at an angle B = 30° above the horizontal. r=60[m] 0 0=120° LAUNCH POINT ski jumper B -GROUND Figure 2: Ski jumper attempting a ski jump. Part (a) [8 marks] Assuming a stationary start from 0 = 0 , show that, at launch (when e = 120°): 1. [4 marks] The velocity of the ski jumper is 293 [m/s]. 11. [4 marks] The normal reaction acting on the ski jumper is 1657 [N]. Part (b) [2 marks] Calculate the height of the launch point above the ground. Part (c) [8 marks] From parts (a) and (b), determine: 1. [4 marks] The horizontal distance (as measured from the launch point) travelled by the ski jumper after launch. You may assume that there is no air resistance. 11. [4 marks] The velocity vector (in x-y coordinates) of the ski jumper just before impacting the ground.

Answer 1

-Gom سم h₂ 120 Kinetic energy Ground ref. (a) Pt 030°, the ski-Jumper start from rest, so no (il using conservation of energy from start to Launch pt- K'El + P.Ei= KE2+ P.E2 high to = 12 w²t migh → 9•81X60 = V2+ 9.81*(60-301) h = 60-6000s 30° 62 = 2x9.81_3053] vr32 mis 60 - 20-3053 (ii) Normal Reactions clearly, mg mu (Centrifugal force] (due tu curved monos N=muz + mg cos300 R = 765x(32)? + 65x9. 6sxa-81x84] bo Nx = 1661.55 [b] Height of Launch point above ground Bil 60 h = 60-60cos30 3013 = 8.038m 60- CS

[c]. obter Launch : y=320m/s (1) Same as profechle motion velter Launch 8-32ms at an angle = 30° ΑΙ 03 930- x L ?? Use = see Tf 32 9-8/ Vy= 325m 30" 32 Cos 30° ylture) c foom A to B) = 20 smo 2x32 X 51230 g 9.81 Time taken to travel th= 8.03 verncelly downward from Ite. Se ut +3074 > 8.03 = 32x4 xt ++*9-81842 = 8.03 = 16+ + 4. gost² solving this quadratic eyn; t = 0.442 sec 0.442 x 32x B ors Ra-Cucongt 9-81 २ = 102.64 m from Launch pt (1) velocity vector just abter Impacting the ground? Vy = (3241030") *0.642 + 9-818 0-442) [u=urgt = 11.408 mis V2-re cos30' - 32х3, ☺67.712 i + 11.4089) m net point c. = 1653 = 27. 7/27A

B [d]. W vertical Impact over landing occurs over Oos see ie. Dt=0.55 Mx = 0.2 Fot = (mV2-rov) Uy = 11.908 m/s Let us assume perfectly inelaste cousion: ie. Ių=0 FN = (mv) 65X 11.608 0.5 = 1483.040V u= 32Cos 30' = 2653 mis at time of Landing >= 1653 mis s MN 기 Let s be the cisternce pavelled and I be the time ie=utat o=1653- (gxt ta 1613 = 19.124 sce 0.2x 9.81 Now (6) Now Horizontal distance travelled S = 1655x14.124 0-289-81 x (14.92472 2 = 195.710m
I have mentioned each steps in solving the problem and also i have used different equations of motion and energy conservation.Verify once if any calculation mistake.

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