(Here I am making an assumption that air resistance is negligible and that the height of the building is negligible compared to the radius of the Earth)
Using SUVAT equations,
#v=u+at#
where #v# is the velocity of the object as a function of time, #u# is the initial velocity at #t =0#, #a# is the acceleration of the object, and #t# is any instant of time. As the object is dropped from a building, the horizontal component of its velocity will always remain 0, i.e. #v_x=0#, hence the vertical component will equal the velocity of the object, i.e. #v_x=v#; also, as it is dropped from a building, the vertical component of its initial velocity is 0, i.e. #u_y=0#.
As it is dropped from the building, the resultant force acting on the object is only the gravitational force acting on it due to Earth's gravitational field. The acceleration #a# of the object is hence equal to acceleration due to gravity #g, approx 9.81ms^-2#
Therefore, #v=v_y=u_y+ g t approx 0+ 9.81t#
#v approx 9.81t#
Depending on the direction you choose, #v# and #g# can be positive or negative
