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

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SUMMARY

The discussion focuses on calculating the energy lost due to air drag in ski jumping. A skier with a mass of 49 kg leaves the ramp at a velocity of 28 m/s at an angle of 22 degrees but lands at a speed of 17 m/s. The correct method to determine the energy lost involves calculating the kinetic energy (KE) both with and without air drag, rather than finding the difference in velocities. The accurate approach reveals that the energy lost due to air resistance is derived from the difference in kinetic energies, not from velocity differences.

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  • Understanding of kinetic energy (KE) calculations
  • Basic physics concepts related to motion and forces
  • Familiarity with air resistance effects on moving objects
  • Ability to perform calculations involving mass and velocity
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  • Study the principles of kinetic energy and its formula: KE = 1/2 mv²
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Physics students, sports scientists, and engineers interested in the dynamics of ski jumping and the effects of air drag on performance.

ViewtifulBeau
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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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