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Help with this physics problem!?

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A 0.60 kg basketball is dropped out of the window that is 6.3 m above the ground. The ball is caught by a person whose hands are 1.3 m above the ground. How much work is done on the ball by its weight?



What is the gravitational potential energy of the basketball, relative to the ground when it is released?

What is the gravitational potential energy of the basketball when it is caught?

How is the change (PEf - PE0) in the ball's gravitational potential energy related to the work done by its weight?

asked Jan 12, 2013 in PHYSICS by anonymous Apprentice

1 Answer

0 votes

potential energy P = mgh, m = mass, g = gravitational constant, h = height

The gravitational potential energy of the basketball, relative to the ground when it is released:

h = 6.3 m

p = 0.6 * 9.8 * 6.3 = 37.044 kg·m2/s2 = 37.044 J

What is the gravitational potential energy of the basketball when it is caught:

h = 6.3 - 1.3 = 5 m

p = 0.6 * 9.8 * 5 = 29.4 kg·m2/s2 = 29.4 J

Source: http://answers.yahoo.com/question/index?qid=20111205170021AAEnp3c

answered Jan 12, 2013 by steve Scholar

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