Convert the initial velocity to meters per second.

The initial velocity $u = 36 \text{ km h}^{-1}$.

$$u = 36 \text{ km h}^{-1} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 10 \text{ m s}^{-1}$$

The driver takes $0.5 \text{ s}$ to react.

$$d_{\text{reaction}} = u \times t_{\text{reaction}}$$ $$d_{\text{reaction}} = 10 \text{ m s}^{-1} \times 0.5 \text{ s} = 5 \text{ m}$$

The initial velocity for braking is $u = 10 \text{ m s}^{-1}$, the final velocity $v = 0 \text{ m s}^{-1}$ (since the bus stops), and the acceleration $a = -2.5 \text{ m s}^{-2}$ (negative because it's deceleration).
Using the kinematic equation $v^2 = u^2 + 2as$:

$$0^2 = (10 \text{ m s}^{-1})^2 + 2(-2.5 \text{ m s}^{-2})s_{\text{braking}}$$ $$0 = 100 \text{ m}^2 \text{ s}^{-2} - 5 \text{ m s}^{-2} \cdot s_{\text{braking}}$$ $$5 \text{ m s}^{-2} \cdot s_{\text{braking}} = 100 \text{ m}^2 \text{ s}^{-2}$$ $$s_{\text{braking}} = \frac{100 \text{ m}^2 \text{ s}^{-2}}{5 \text{ m s}^{-2}} = 20 \text{ m}$$

The total stopping distance is the sum of the reaction distance and the braking distance.

$$D_{\text{total}} = d_{\text{reaction}} + s_{\text{braking}}$$ $$D_{\text{total}} = 5 \text{ m} + 20 \text{ m} = 25 \text{ m}$$

The obstacle is $30 \text{ m}$ ahead.
Since $25 \text{ m} < 30 \text{ m}$, the bus will stop before reaching the obstacle.

The bus will be able to stop before reaching the obstacle.
The total stopping distance is $\text{25 m}$.