The figure presents a skier at Point A, at the top of a ski jump, which...

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Physics

The figure presents a skier at Point A, at the top of a skijump, which curves steeply downward and to the right, passesthrough point B, and then curves back upward and ends at point C.The height at point A is h sub 1, the height at point B is 0meters, and the height at point C is 10 meters. h sub 1 is greaterthan 10 meters. A ski jumper starts from rest from point A at thetop of a hill that is a height h1 above point B at the bottom ofthe hill. The skier and skis have a combined mass of 80 kg. Theskier slides down the hill and then up a ramp and is launched intothe air at point C that is a height of 10 m above the ground. Theskier reaches point C traveling at 42ms. (a) Calculate the heighth1. Question 2 (b) Calculate the speed of the skier as the skierreaches point B. Question 3 (c) Is the work done by thegravitational force on the skier as the skier slides from point Ato point B positive or negative? ____ Positive ____ NegativeJustify your answer. 4 The skier leaves the ramp at point Ctraveling at an angle of 25° above the horizontal. (d) Calculatethe kinetic energy of the skier at the highest point in the skier'strajectory. Question 5 (e) i. Calculate the horizontal distancefrom the point directly below C to where the skier lands. Question6 ii. If the angle is increased to 35°, will the new horizontaldistance traveled by the skier be greater than, less than, or equalto the answer from part (e)(i) ? ____ Greater than ____ Less than____ Equal to Justify your answer. After landing, the skier slidesalong horizontal ground before coming to a stop. The skier’sinitial speed on the ground is the horizontal component of theskier’s velocity when the skier left the ramp. The averagecoefficient of friction μ is given as a function of the distance xmoved by the skier by the equation μ=0.20x. (f) Calculate thedistance the skier moves between landing and coming to a stop.

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3.7 Ratings (512 Votes)
height at A h1 height at C 10 m change in Pot energy from A to C mg h110 KE at C 12 mv2 starts at A from rest gain in KE at C change in Pot energy from A to C mg h110 12 mv2 v 42 ms speed at C a h1 v22g 10 422298 10    See Answer
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