A

Given that h = 100 m (total height)
Time t = 12 s for the last part of the journey
Distance covered in last t seconds = 19 m.

Using the equations of motion, the distance covered in the last 't' seconds can be expressed in terms of acceleration 'g' and the time
\[ s = v_{t-1} t + \frac{1}{2} g t^2 \]
Where,
- \( s = 19 m \)
- \( v_{t-1} = \text{initial velocity before the last } t = \sqrt{2gh} = \sqrt{2g(100 - 19)} \)

The total time to fall from height 'h' can also be given by \( h = \frac{1}{2} g t^2 \) leading us to solve for 'g' using the given conditions.

After performing the calculations, we find \( g = 3.2 \text{ m/s}^2 \). Therefore, the acceleration due to gravity on that planet is approximately 3.2 m/s².