Question

A 90 kg skier at the top of a 400 m icy hill slides down, through a 20 m patch of snow (µk= 0.25) and hits a 20 m long jump ramp at a 45° incline. a) (10 points) How far from the edge of the ramp does the skier land (m)? b) (5 points) How high will he get of the ground (max height (m)) ?

Ho 400 m 20 m 45° X 20 m

Answer 1

Use conservation of energy, yagh h = 1 mua - Sagh 2x9.8 x 400 88.544 mis so skier Skier will have velocity OF 88.544 m/s at the foot of the bill. icy Now, work done by friction change in Kinetic energy => -allk' X 2om. Cv-v3) 2 => -0.259.8 x 20 2 Vf- 88.544 => Vf=87.9886 m/s so I at the base of long jump ramp Skier's be 87.9886 m/s velocity will now, energy, + again using conservation of Anh V2 = nugh (87.9886) hanno via 2 = 9.88 20 Sin 450 2 => v' = 86.399 m/s

So, Skier will leave the ramp with 86-399 mis Now, we will use calculate range projectile motion concept to and height. in y- direction y = yo + Voyt - g +² => 0 = 20 sin 45° + 86.399 Sin 450 t 1 x 9.8 t² 4.9t- 61.0933 t 14.1421 t solving, 12.6953 sec in x-direction 2 86.399 x cos 450 12.6953 775.6m. a) so skier will and 775.6 m. from the edge of the ramp b) y = Yo + voyt 4 g +² 20 sin 45° + - y 86.399 Sin 450 t - *9.8ť? mazimum height Vertical component of Velocity is zero, now, at

soj ve every -go ت ts Voy 86.399 X Sin450 6.234 S 7 9.8 oo Sol 204.571m so Skier will 204.571m. of the ground, get high