Question

Consider 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, is defined from the horizontal as shown. The ski jumper goes along the ski jump track for 0 = 120° before launching at an angle B = 30° above the horizontal. r=60[m] 0 0=120° LAUNCH POINT ski jumper yo GROUND Figure 2: Ski jumper attempting a ski jump.

Part (a) [8 marks]

Assuming a stationary start from θ = 0°, show that, at launch (when θ = 120°):

1. [4 marks] The velocity of the ski jumper is ≈93 [m/s].
2. [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.
2. [4 marks] The velocity vector (in x-y coordinates) of the ski jumper just before impacting the ground.

Part (d) [12 marks]

We now consider the kinetics of the ski jumper after landing. The vertical impact upon landing occurs over 0.5 [s], and the coefficient of (kinetic) friction, μ between the skis and snow is 0.2. Determine:

1. [4 marks] The total normal force exerted by the ground on the ski jumper during vertical impact.
2. [4 marks] The time taken for the ski jumper to come to a horizontal stop after landing.
3. [4 marks] The horizontal distance covered by the ski jumper before coming to a stop.

Answer 1

hey your answer of 93m/s of velocity is not possible may that is wrong , here is the actual solution

65kg M=lom Reference datum ote 0=1260 30° ski Jumper t 3 = 30° ng Dific Part @ P. Fico KE:0 P. Es = -mg de cos 30º =-33133.2665 KEF muz theorem. P. Eit Ki Eig DEST K. Ef work- energy - 33133.266 + ļmuz 31.929 mis Uz Normal цаснщ at launch Poll N N-mg cos30° Tom re you mg cossot r 65C 9.81% (0930'+ 31.9242 Go (IN IV. 1657N 31.9122232 1656-637 N Part 6 Height of laemch Point h = R- Rosso = 60- 6o cos30 8.0385m Parto first calculate time of sught ť in vertical dir 7 EUISA Ź 9t2 -8.0395= 15.96454 4.90572 4.905t2_15.964st -9.0385 = -(-15-9645) + J(15.9645) + 4x8.0385x 4.905 2x 4.905 t: 3.6983

Now Horinental distance U cos 30t 31.929* C0530x 3.648 102.255 m Velocity vector V = vcoso? - esind? VE vit at in xadir" U cos p Vosa = 27.651 m/s in y-dut vina Usinp 9.81t 15. 9645- 9.81 x 3.698 Vsina - - 20.3129 <i 27.651 Î – 20.3129 ſ Parta NIE 27-6372+ 20.31282 34.310 mis FV L = tem tem-l 20.3129 27-651 ) mg 22 in y-dii 36-302 final momentum - Initial momento Fri Ev (ver ertical impact 65 x 20.3129 force)= dla vy) -+2640-677N 0.5 N = mg -F = 65 x 9.81 2640.677 Na - 2003.027 Normal exerted ground, by

fk= un, - 0.2 x 2003-027 = 400. 6054 N ax fk -400-6054 - Go 16316 m/s2 65 84 skis is deceleratiay When Stops V20 Vcosa + cxt ta 27.651 4.48658 to stop こ 6.16316 distance covered d = vosdot 97-651x4.4965 + Zaxt2 5x6-16316 x 14.48652 Td = 62.029 m