Answer:
The final velocity, [tex]v_{f}[/tex], is gd.
Explanation:
The condition here is a free falling object. Thus from the third equation of motion under free fall, we have;
[tex]v_{f}[/tex] = [tex]v_{i}[/tex] + 2gs
where [tex]v_{f}[/tex] is the final velocity of the object, [tex]v_{i}[/tex] is the initial velocity of the object, g is the gravitational force and s is the height.
Since the object falls from a height of d, then [tex]v_{i}[/tex] = 0 m/s, and s = d.
So that;
[tex]v_{f}[/tex] = 0 + 2gd
= 2gd
[tex]v_{f}[/tex] = 2gd
When the distance is [tex]\frac{d}{2}[/tex], [tex]v_{i}[/tex] = 0 m/s.
Then;
[tex]v_{f}[/tex] = 2g[tex]\frac{d}{2}[/tex]
[tex]v_{f}[/tex] = g x d
When the object falls through the height [tex]\frac{d}{2}[/tex], then the final velocity is gd.
