Let's return to our ball-Earth example, only now let's examine a case where a ball is rising in the air. You toss a ball with a mass of 0.703 kg upward. After it leaves your hand, it continues to rise while slowing down (before eventually reversing direction and falling again). Let's say it rises a distance of 1.30 m above your hand (its original height).   Does the gravitational potential energy of the ball-Earth system increase, decrease, or stay the same?   Calculate the change in gravitational potential energy (in J). (Be sure to include the correct sign.)  ________J

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Chapter1: Units, Trigonometry. And Vectors
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We are now able to define a mathematical formula for gravitational potential energy.
Near the Earth's surface, the gravitational potential energy of a system consisting of the earth and an object with a mass m is
EP = mgh,
where g is the acceleration of gravity (9.80 m/s2) and h is the height above ground level (positive upward).
Note that the "ground level" could really be any height we choose, because what's really important is the change in potential energy. The difference between two heights always gives the same change in potential energy, regardless of where we set the "zero" of height. In other words, if we find the change in potential energy 
ΔEP = EP,f − EP,i,
 the final potential energy minus the initial, we have
ΔEP = mghf − mghi = mgΔh.
The change in gravitational potential energy is just mg times the change in height.
Let's return to our ball-Earth example, only now let's examine a case where a ball is rising in the air.
You toss a ball with a mass of 0.703 kg upward. After it leaves your hand, it continues to rise while slowing down (before eventually reversing direction and falling again). Let's say it rises a distance of 1.30 m above your hand (its original height).
 
Does the gravitational potential energy of the ball-Earth system increase, decrease, or stay the same?
 
Calculate the change in gravitational potential energy (in J). (Be sure to include the correct sign.)
 ________J
 
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