a) A 72 kg skier leaves the end of a ski-jump ramp with a velocity of 20 m/s dir

Question

a) A 72 kg skier leaves the end of a ski-jump ramp with a velocity of 20 m/s directed 25° above the horizontal. Suppose that as a result of air drag the skier returns to the ground with a speed of 18 m/s, landing 13 m vertically below the end of the ramp. From the launch to the return to the ground, by how much is the mechanical energy of the skier-Earth system reduced because of air drag? ___ J (3 sig figs)

b) For a three-particle system, with masses m1 = 2.1 kg @ (0,0) , m2 = 4.0 kg @ (2,1) and m3 = 9.0 kg @ (1,2)

What are the coordinates of the center of mass?
___ m (x-coordinate)
___ m (y-coordinate)

Explanation / Answer

(a) Mass m = 72 Kg Initial speed u = 20 m/s Angle of projection ? = 25 degrees Hight H = 13 m Let v be the final speed of the ball while touching the ground. vx = u cos25 = 18.126 m/s (vy)^2 = (u sin25)^2 + 2 (-g)(-H)           = 71.44 + 254.8             = 326.24 vy = 18.06 m/s Actual Final speed v = Sqrt[(18.126)^2 + (18.06)^2]                      = 25.58 m/s Final speed v' = 18 m/s Mechanical energy loss in Earth & skier system due to air drag = Work done                      = (1/2)m [(25.58)^2 - (18)^2]                        = 0.5 * 72 * 330.71                      = 1.19*10^4 J (b) m1 = 2.1 Kg ; x1 = 0 , y1 = 0 m2 = 4 Kg     ; x2 = 2 , y2 = 1 m3 = 9 Kg     ; x3 = 1 , y3 = 2 Coordinates of the center of mass Xcm = (m1x1 + m2x2 + m3x3)/(m1 + m2 + m3)        = (0+8+9)/(2.1+4+9) = 17/15.1 = 1.125 Ycm = (m1y1 + m2y2 + m3y3)/(m1 + m2 + m3)        = (0+4+18)/(2.1+4+9) = 22/15.1 = 1.456

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