Gravitational falling and kinematics

Question:
A stunt man jumps off from a 202-m high building onto a cushion having a thickness of 2.00 m. After his going into the cushion, it is crushed to a thickness of 0.500 m. What is the acceleration as he slows down?

Answer:
We need to find his velocity right before reaching the cushion. The cushion has 2.00-m high, so the stunt man falls by 200 m. The displacement, the initial velocity (0 m/s) and the acceleration (-9.80 m/s$^2$) are known. In order to find the final velocity, we use
\begin{eqnarray}
v^2_f - v^2_0 &=& 2g\Delta h  \\
v_f &=& \sqrt{2g\Delta h}  \\
      &=& \sqrt{2 \cdot (-9.8) \cdot (-200)}
     &=& 62.61 \ \mathrm{m/s}
\end{eqnarray}
Now, we calculate the deceleration of the man by the cushion using the same equation. The final velocity is zero and the initial velocity is 62.61 m/s. The final thickness is (0.500-2.00) m. Thus,
\begin{eqnarray}
v^2_f - v^2_0 &=& 2ad  \\
0 - 62.61^2 &=& 2a (0.5 - 2)  \\
a &=& \frac{-62.61^2}{2 \cdot (-1.5)}  \\
a &=& 1307 \ \mathrm{m/s}^2
\end{eqnarray}

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