How Much Energy Is Lost Due to Air Drag in Ski Jumping?

AI Thread Summary
The discussion focuses on calculating the energy lost due to air drag in ski jumping, specifically for a skier who leaves the ramp at 28 m/s and lands at 17 m/s. The initial calculations incorrectly used the difference in velocity to find kinetic energy (KE), resulting in an erroneous energy loss of 5015 J. The correct approach involves calculating the KE for both the initial and final velocities and then finding the difference between these two values. This is crucial because KE is a square function of velocity, making the initial method flawed. The correct calculation method is emphasized for accurate results in determining energy loss due to air resistance.
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A 49kg skier leaves the end of a ski-jump ramp with a velocity of 28 m/s directed 22o above the horizantal. Suppose that as a result of air drag the skier returns to the ground with a speed of 17 m/s, landing 10 meters vertically below 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?

first i found the velocity the skier should have without air drag (31.3075 m/s). then i found the difference between 31.3075 and the actual speed 17m/s. then i used that number and plugged it into 1/2 m v^2 = KE to find the energy lost due to air resistance. 5015 J... this isn't right though. what did i do wrong?
 
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ViewtifulBeau said:
first i found the velocity the skier should have without air drag (31.3075 m/s). then i found the difference between 31.3075 and the actual speed 17m/s. then i used that number and plugged it into 1/2 m v^2 = KE to find the energy lost due to air resistance. 5015 J... this isn't right though. what did i do wrong?

You shouldn't calculate the difference in velocity and then the corresponding hypothetical KE, you should take the difference between the 2 KE (with and without drag).

cheers,
Patrick.
 
Why: because KE is a square function of velocity, and therefore finding the difference in velocity and plugging that into the KE equation will give the wrong answer.
 
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