When they were falling, their acceleration is equal to the gravity acceleration.
When they started stopping, the time required to stop is 1.5 seconds, that is, 3 times less than the time accelerating.
Since the change in velocity when falling and when stopping is the same (same magnitude but opposite signal), the acceleration when stopping will be 3 times more than the acceleration when falling (because the change in velocity is the same and the time is 3 times less).
So we have:
[tex]\begin{gathered} a=-3g\\ \\ a=-3\cdot(-9.8)\\ \\ a=29.4\text{ m/s^^b2} \end{gathered}[/tex]
Therefore the acceleration is 29.4 m/s².
