$$v^2 = u^2 + 2as$$
where:
* v is the final velocity of the object (in m/s)
* u is the initial velocity of the object (in m/s)
* a is the acceleration due to gravity (in m/s²)
* s is the distance the object has fallen (in m)
In this case, the initial velocity of the object is 0 m/s, the acceleration due to gravity is -9.8 m/s², and the distance the object has fallen is 120.0 m. Substituting these values into the equation, we get:
$$v^2 = 0 + 2(-9.8)(120.0)$$
$$v^2 = -2352.0$$
Taking the square root of both sides, we get:
$$v = \sqrt{-2352.0}$$
$$v = 48.5 \text{ m/s}$$
Therefore, the velocity of the object when it hits the ground is 48.5 m/s.
