Well I get:
initial KE = (1/2)(60)(26)^2 = 20,280 J
gain due to fall = m g h = 60(9.81)(8.3) = 4485 J
so total energy at ground if no friction = 25,165 J
actual Ke at ground = (1/2)(60)(19)^2 = 10,830 J
difference = 14,335 J which is the loss due to air drag
A 60 kg skier leaves the end of a ski-jump ramp with a velocity of 26 m/s directed 23° above the horizontal. Suppose that as a result of air drag the skier returns to the ground with a speed of 19 m/s, landing 8.3 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?
I got -14330.4J, but this answer is wrong and I can't figure out why
I did KE= 09450
PE=-4880.4
ME=-14330.4
thanks
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