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When the ball leave your hand it goes up with an initial velocity v0 so that the action of your hand gives the ball the kinetic energy

K=0.5 M v02

where M is the ball mass.

When the ball goes up the kinetic energy decreases: a part is converted in potential energy while the balls is higher and higher and part is dissipates due to the attrition with air (that more precisely is due to air viscosity) and to the fact that air particles are put in motion by the arrival of the ball so that they gain kinetic energy at the expenses of the kinetic energy of the ball due to collision.

Neglecting the last term that, due to the small density of air with respect to the ball is generally quite small, the viscosity dissipate energy mainly by generating heat. When the ball stops at maximum height the kinetic energy is reduced to zero (the velocity is zero) and all the energy is potential, while a certain amount of heat has been dissipated during the motion.

If we call Q the quantity of generated heat and h the maximum height reached by the ball, for the energy conservation rule we have

0.5 M v02=Q+M g h

If the air viscosity is negligible, or if the ball goes up in vacuum, Q is zero and we can deduce the maximum height the ball reaches

h=(M v02)/( 2 M g)

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