Verify that,
Final Velocity(v) = √2gh​

The equation gives the velocity of a falling object just before it hits the ground - a consequence of the conservation of energy (the energy of the object before being released is its gravitational potential energy; the energy of the object just prior to hitting the Earth is its kinetic energy, and they are equal).

The equation however assumes no air resistance, which on Earth (or on any large body with an atmosphere) isn’t exactly correct. Also, gravitational field strength varies with height.

mgh=12mv2+12ρACDhv2

⟹v=2mghm+ACDhρ−−−−−−−√  - this incorporates the affects of drag on the motion of the object. An even more accurate answer for this would involve integrating the expression for drag force with respect to height to obtain the work done by frictional forces.

Explanation:

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