Conservation of energy of ball with drag

AI Thread Summary
A .63 kg ball is thrown upward at 14 m/s, reaching a height of 8.1 m, and the energy dissipated by air drag during ascent is being calculated. The kinetic energy (KE) is computed as 61.74 J using the formula KE = 1/2 MV^2. The potential energy (PE) at maximum height is determined using the formula PE = mgh. The difference between the initial kinetic energy and the potential energy reveals a loss of 12 J due to air drag. This analysis highlights the impact of drag on energy conservation in projectile motion.
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Homework Statement


A .63 kg ball is thrown up with an initial speed of 14 m/s and reaches a maximum height of 8.1m. How much energy is dissipated by the air drag acting on the ball during the ascent?


Homework Equations


KE= 1/2MV^2


The Attempt at a Solution


I'm not very good at physics. I think I have to find the joules by plugging in the numbers accordingly in the equation, but I don't know what to do from there...

KE = .5(.63)(14)^2 = 61.74

The actual answer is -12 J >.<
 
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Compare the change in kinetic energy with the change in potential energy. Which is mgh. The difference is the loss to friction.
 

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