Question

The figure presents a skier at Point A, at the top of a ski jump, which curves steeply downward and to the right, passes through 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 0 meters, and the height at point C is 10 meters. h sub 1 is greater than 10 meters. A ski jumper starts from rest from point A at the top of a hill that is a height h1 above point B at the bottom of the hill. The skier and skis have a combined mass of 80 kg. The skier slides down the hill and then up a ramp and is launched into the air at point C that is a height of 10 m above the ground. The skier reaches point C traveling at 42ms. (a) Calculate the height h1. Question 2 (b) Calculate the speed of the skier as the skier reaches point B. Question 3 (c) Is the work done by the gravitational force on the skier as the skier slides from point A to point B positive or negative? ____ Positive ____ Negative Justify your answer. 4 The skier leaves the ramp at point C traveling at an angle of 25° above the horizontal. (d) Calculate the kinetic energy of the skier at the highest point in the skier's trajectory. Question 5 (e) i. Calculate the horizontal distance from the point directly below C to where the skier lands. Question 6 ii. If the angle is increased to 35°, will the new horizontal distance traveled by the skier be greater than, less than, or equal to the answer from part (e)(i) ? ____ Greater than ____ Less than ____ Equal to Justify your answer. After landing, the skier slides along horizontal ground before coming to a stop. The skier’s initial speed on the ground is the horizontal component of the skier’s velocity when the skier left the ramp. The average coefficient of friction μ is given as a function of the distance x moved by the skier by the equation μ=0.20x. (f) Calculate the distance the skier moves between landing and coming to a stop.

Answer 1

height at A = h₁

height at C = 10 m.

change in Pot. energy from A to C = mg( h₁-10)

KE at C = 1/2 mv² ,

starts at A from rest , gain in KE at C = change in Pot. energy from A to C

mg( h₁-10) = 1/2 mv²

v = 42 m/s speed at C

a) h₁ = v²/2g +10 = 42²/2*9.8 + 10 = 100 m

height at B =0

b) speed at B v = sqrt(2gh₁) = sqrt( 2*9.8*100) = 44.27 m/s

c) work done is +ve , results in +ve KE of the skier

d) highest point of trajectory is A and speed =0 , KE =0

e) height of C = 10 m

time of fall t= (2gh)^1/2 = sqrt( 2*9.8*10) = 14 s

The skier falls vertically under gravity , initial vertical vel. =0

The skier moves horizontally from point C at 42 m/s  

horizontal distance moved = 42 * 14 = 588 m , lands at 588 m from C

f) launch angle = 35 deg.

horizontal component of vel. = 42* Cos(35) = 34.4 m/s

vertical comp . = 42*Sin(35) = 24.09 m/s

The skier will reach a maximum height above C and then descend to ground.

time to reach max. height = v_yvertical /g = 24.09 /9.8 = 2.46 s

height from C h = v_ver²/2g = 24.09²/ 2*9.8 = 29.61 m

Now the skier has to descend 29.61 +10 = 39.61 m to reach the ground

time to descend t = sqrt( 2*9.8*39.61) = 27.86 s

Total time of flight = 28.86+ 2.46 = 31.32 s

Total horizontal distance = 34.4 * 31.32 = 1077.4 m

greater than the previous case.