The air molecules rushing out of the balloon represents a gain in (downward) momentum. The rate at which this momentum is being generated is equal to the downward force the balloon is exerting to eject the air molecules. By N3L, this is the upward thrust exerting on the balloon.
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When you’re hit by a water cannon, you’re continuously bombarded by godzillion number of H2O molecules at any one instant. The force exerted by each molecules is tiny and fleeting (lasts only for an instant). But collectively, they exert a large and continuous force.
To calculate the force exerted by a water cannon, we turn to calculating the force experienced by the water cannon instead. When the water hits you, it loses momentum. The rate at which it is losing (or changing) momentum is equal to the force that you are exerting on the water. By N3L, the water is exerting an equal but opposite force on you. Ouch.
xmPhysics 