To find the height to which the lift takes the passenger, we need to calculate the displacement. The displacement is given by the area under the speed-time (v-t) graph .

Based on the standard graph associated with this specific problem (often found in competitive exams like NEET/AIPMT 2013), the motion corresponds to a trapezium with the following phases:

1. Acceleration: From $t = 0$ to $t = 2$ s, speed increases from 0 to 3.6 m/s.
2. Constant Speed: From $t = 2$ to $t = 10$ s, speed remains constant at 3.6 m/s.
3. Deceleration: From $t = 10$ to $t = 12$ s, speed decreases from 3.6 m/s to 0.

Calculation: $\text{Height} = \text{Area under the curve} = \text{Area of Trapezium}$ $\text{Area} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times (\text{Height})$ Sum of parallel sides = Total time + Time of constant motion = $12 \text{ s} + (10 - 2) \text{ s} = 12 + 8 = 20 \text{ s}$. Height of graph = Maximum speed = $3.6 \text{ m/s}$.

$\text{Height} = \frac{1}{2} \times (20) \times 3.6$ $\text{Height} = 10 \times 3.6 = 36 \text{ meters}$

The mass of the lift (1500 kg) is not required for calculating the displacement from the v-t graph.