Here is a classic textbook application of Newton’s Laws. The analysis of forces and torques on the two masses and pulley gives:

$$\begin{align} m_{2} g − T_2 &= m_2 a \\ T_1 − m_1 g &= m_1 a \\ R(T_2 − T_1 ) &= I\alpha \notag \\ &= I \frac{a}{R} \notag \\ &= \beta m_p R^2 \frac{a}{R} \notag \\ R(T_2 − T_1 ) &= \beta m_p R a \end{align}$$

Solve Equation (1), (2) and (3) to obtain the system’s acceleration:

$$\begin{align} a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \beta m_p} \tag{4} \end{align}$$

Note: Equation (4) is the net force on the system divided by its total mass, where $\beta m_p$ is the effective rotational inertia of the pulley.

For constant acceleration the time to fall a distance h from rest is:

$$\begin{align} t=\sqrt{\frac{2h}{a}}\tag{5} \end{align}$$

In the example where $h =$ 100 cm, $m_1 =$ 500g, $m_2 =$ 550g, $β =$ 1, $m_p =$ 10g,and $g = 980\text{cm/s}^2$, Equation (4) and (5) give $t =$ 2.1 s.

Since $β m_p << m_1 + m_2$, the effect of the pulley may be neglected if desired.