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:
\(mg \Delta h = \frac{mv^2}{2}\)
Now, we multiply by \(\frac{2}{m}\) to get:
\(v^2=2g \Delta h\)
Then, we have that:
\(v=\sqrt{2g \Delta h}\)
Now, we substitute. \(g=9.8 \,m/s^2\) while \(\Delta h = 30 \,m\). Therefore,
\(v = \sqrt{2g \Delta h} = \sqrt{588\, m^2/s^2} = ~24.2 \, m/s\).
