Answer:
Approximately [tex]7.0\; \rm m \cdot s^{-1}[/tex].
Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.
Explanation:
Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant [tex]g = 9.81 \; \rm m \cdot s^{-2}[/tex] near the surface of the earth.
For an object that is accelerating constantly,
[tex]v^2 - u^2 = 2\, a \, x[/tex],
where
In this case, [tex]x[/tex] is the same as the change in the ball's height: [tex]x = 2.5\; \rm m[/tex]. By assumption, this ball was dropped with no initial velocity. As a result, [tex]u = 0[/tex]. Since the ball is accelerating due to gravity, [tex]a = 9.81\; \rm m \cdot s^{-2}[/tex].
[tex]v^2 - u^2 = 2\, g \cdot h[/tex].
In this case, [tex]v[/tex] would be the velocity of the ball just before it hits the ground. Solve for
[tex]v^2 = 2\, a\, x + u^2[/tex].
[tex]\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}[/tex].
