A 2.5 kg object is dropped from a height of 10.0 m above the ground, what is its speed when it hits the ground (ignore air resistance)?

What we are trying to find

The object starts from rest and falls straight down, so we want its final speed after dropping a vertical distance of $10.0\,\text{m}$ under gravity (no air resistance).

Using a kinematics relation for free fall

For constant acceleration, one helpful equation is: $$v^2 = v_0^2 + 2a\Delta y$$ Here, $v_0=0$ (dropped), $a=g=9.8\,\text{m/s}^2$, and $\Delta y = 10.0\,\text{m}$ downward.

Plug in values and solve

$$v^2 = 0^2 + 2(9.8)(10.0) = 196$$ $$v = \sqrt{196} = 14.0\,\text{m/s}$$

Quick check and interpretation

The units work out because $\text{m/s}^2 \cdot \text{m} = \text{m}^2/\text{s}^2$, and taking the square root gives $\text{m/s}$. The given mass $2.5\,\text{kg}$ is extra information here; in ideal free fall, the speed depends on $g$ and the drop height, not the mass.