We can solve this using the principle of conservation of mechanical energy, as there are no non-conservative forces like friction. The system consists of the man, the sandbag, and the Earth. Let the ground be the zero potential energy level ($U=0$).

The initial energy of the system ($E_i$), when the man is at height $h$ and everything is at rest, is purely potential energy:

$$E_i = K_i + U_i = 0 + m_1gh + m_2g(0) = m_1gh$$

The final energy of the system ($E_f$), when the man hits the ground and the sandbag is at height $h$, has both kinetic and potential energy. Both objects move with the same speed $v$.

$$E_f = K_f + U_f = \frac{1}{2}(m_1+m_2)v^2 + m_1g(0) + m_2gh = \frac{1}{2}(m_1+m_2)v^2 + m_2gh$$

By conservation of energy, $E_i = E_f$:

$$m_1gh = \frac{1}{2}(m_1+m_2)v^2 + m_2gh$$

Now, we solve for the final speed $v$:

$$(m_1-m_2)gh = \frac{1}{2}(m_1+m_2)v^2$$ $$v^2 = \frac{2(m_1-m_2)gh}{m_1+m_2}$$