How High Does a Basketball Bounce After Losing 20% Energy?

AI Thread Summary
A basketball dropped from a height of 2 meters loses 20% of its energy due to air friction before bouncing. The initial potential energy (PE) is calculated using the formula PE = mgh, where g is 9.8 m/s². The total energy before the drop equals the total energy after the bounce, leading to the equation mgh1 = 0.80(mgh2). The challenge lies in determining the height of the bounce (h2) without knowing the mass or kinetic energy. The discussion highlights the need to clarify the initial and final kinetic energy values to solve for the bounce height accurately.
aleferesco
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Homework Statement



A basketball is dropped from a height of 2m. It lost lost 20% of its energy, in the bouncing from the floor due to friction in the air, before it reaches the highest point of its first bouncing. How high did it bounce?

h1= 2m
h2=?
g=9.8m/s^2

Homework Equations



PE = mgh and Energy Total (initial)=Energy Total (final)

mgh1 = 0.20 + mgh2

The Attempt at a Solution



I don't know how to solve it without the mass and kinetic energy
 
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aleferesco said:

Homework Statement



A basketball is dropped from a height of 2m. It lost lost 20% of its energy, in the bouncing from the floor due to friction in the air, before it reaches the highest point of its first bouncing. How high did it bounce?

h1= 2m
h2=?
g=9.8m/s^2

Homework Equations



PE = mgh and Energy Total (initial)=Energy Total (final)

mgh1 = 0.20(?)[/color] + mgh2
check this equation for an error...20% of what?

The Attempt at a Solution



I don't know how to solve it without the mass and kinetic energy
What's the initial KE at the release point? What's the final KE at the top of the first bounce?
 

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