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)?
Ignoring air resistance, the impact speed is found from $v=\sqrt{2gh}$. With $g=9.8\,\text{m/s}^2$ and $h=10.0\,\text{m}$, $v=\sqrt{2(9.8)(10.0)}=\sqrt{196}=14.0\,\text{m/s}$. The mass (2.5 kg) does not affect the final speed in free fall.
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.
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