If you assumed that the drop kept accelerating, then the final velocity is 9.81 * T. The time T is SQR (2*7600 / 9.81) or about 39.4 seconds. The velocity is 9.81 * 39.4 seconds = 386.5 m/s. That is 1391 km/h which is way too fast. So then you have to know what is your model of drag forces. Are you using D = kv or D = kv^2? Drag forces would produce a terminal velocity of the drop well before it got to the speeds above. For example the terminal velocity of a human body is 190 km/h. The terminal velocity of a water drop is going to be well under that value, because its weight is much less than the human body. When the air drag equals the weight there is no downward acceleration and the drop would have reached its terminal velocity. If you have a model for your drag force as a function of v, set it equal to mg (the weight) and solve for the terminal velocity.