We use conservation of energy. Note that at the beginning of the drop, the pig has only potential energy, and at the bottom, it has only kinetic energy. Then:
[tex]mg \Delta h = \frac{mv^2}{2}[/tex]
Now, we multiply by [tex]\frac{2}{m}[/tex] to get:
[tex]v^2=2g \Delta h[/tex]
Then, we have that:
[tex]v=\sqrt{2g \Delta h}[/tex]
Now, we substitute. [tex]g=9.8 \,m/s^2[/tex] while [tex]\Delta h = 30 \,m[/tex]. Therefore,
[tex]v = \sqrt{2g \Delta h} = \sqrt{588\, m^2/s^2} = ~24.2 \, m/s[/tex].
