Answer:
Approximately [tex]9.9\; \rm m\cdot s^{-1}[/tex], assuming that this object started from rest, [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex], and that air resistance on this object is negligible.
Explanation:
If the effect of drag on this object is negligible, the acceleration of this object would be constantly equal to [tex]g[/tex], the gravitational field strength ([tex]g \approx 9.81\; \rm m\cdot s^{-2}[/tex] near the surface of the earth.)
Since acceleration is constant, it would be possible to find the final velocity of this object using SUVAT equations.
The following SUVAT equation would apply:
[tex]v^{2} - u^{2} = 2\, a\, x[/tex].
Rearrange to find an expression for [tex]v[/tex]:
[tex]\begin{aligned}v^{2} &= 2\, a\, x + u^{2}\end{aligned}[/tex].
[tex]\begin{aligned}v &= \sqrt{2\, a\, x + u^{2}}\end{aligned}[/tex].
Substitute in the values [tex]u = 0\; \rm m\cdot s^{-1}[/tex], [tex]a = 9.81\; \rm m\cdot s^{-2}[/tex], and [tex]x = 5\; \rm m[/tex]. Evaluate the expression to find the value of [tex]v[/tex]:
[tex]\begin{aligned}v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 9.81\; \rm m\cdot s^{-2} \times 5\; \rm m} \\ &\approx 9.9\; \rm m\cdot s^{-1}\end{aligned}[/tex].
Therefore, under these assumptions, the velocity of this object would be approximately [tex]9.9\; \rm m\cdot s^{-1}[/tex] right before it hits the ground.
