The drop of water falls from a ledge of a tall building from a height 320 m.

An object initially traveling at speed u m/s and accelerating at a m/s^2, in time t, travels a distance equal to S = u*t + (1/2)*a*t^2. The velocity of the body after time t is equal to v = u + a*t.

When the drop falls, its initial speed is 0. Ignore resistance due to air and take the acceleration due to gravity to be constant and equal to 9.8 m/s^2.

The distance traveled by the drop in the first second is 0*1 + (1/2)*9.8*1 = 4.9 m.

At the start of the 3rd second the speed of the drop is 0 + 9.8*2 = 19.6 m/s. The distance traveled by the drop in the third second is 19.6*1 + (1/2)*9.8*1 = 19.6 + 4.9 = 24.5

The difference between 24.5 and 4.9 is 19.6.

In the 3rd second the drop travels a distance 19.6 m more than it did in the first second after it fell off the ledge.