physics
posted by mary .
A 60 kg skier leaves the end of a skijump 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 skierEarth 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

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
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