A skilled skier knows to jump upward before reaching a downward slope. Consider

Question

A skilled skier knows to jump upward before reaching a downward slope. Consider a jump in which the launch speed is v0 = 10 m/s, the launch angle is 0 = 9.0, the initial course is approximately flat, and the steeper track has a slope of 11.3. Figure (a) shows a prejump that allows the skier to land on the top portion of the steeper track. Figure (b) shows a jump at the edge of the steeper track. In Figure (a), the skier lands at approximately the launch level. (a) In the landing, what is the angle between the skier's path and the slope? In Figure (b), (b) how far below the launch level does the skier land and (c) what is ? (The greater fall and greater can result in loss of control in the landing.)

Explanation / Answer

Vo=10m/s o=9º

a) Landing angle is -9º

b & c)
Vx(skier)=10 cos(9º)
Vy(skier)=10 sin(9º)

X(skier)=10 cos(9º) t
Y(skier)= (-9.8/2) t^2 + 10 sin(9º) t

Y(hill)= tan(-11.3º) x= -0.2 x
Determine Y(hill) as a function of time using X(skier) equation
Y(hill)= -0.2 (10 cos(9º)) t

Skier-Hill intercept condition is Y(skier)=Y(hill)
(-9.8/2) t^2 + 10 sin(9º) t= -0.2 (10 cos(9º)) t

Solving for t yields:
t=0 sec (The takeoff condition.)
t=.722 sec (The downrange intercept condition.)

Y(intercept)= -0.2 (10 cos(9º)) 0.722 = -1.42 meters
dy/dt (skier at intercept)=-9.8 (0.722) + 10 sin(9º)= -5.514 m/s
dx/dt (skier at intercept)=10 cos(9º)=9.876 m/s
(skier at intercept) = atan(Vy/Vx)=atan(-5.514/9.876)= -29.17º
Relative Landing Angle = -11.3º - (-29.17º) = 17.87º (call it 18º)

a) -9º
b) -1.42 m
c) -29.17º (relative to flat course) or 17.87º (relative to hill)

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