Question:

A stunt woman jumps off from a 202-m high building onto a cushion having a thickness of 2.00 m. After her 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 her velocity right before reaching the cushion. The cushion has 2.00-m high, so the stunt woman 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

\[  v^2_f – v^2_0 = 2g\Delta h  \]

\[  v_f = \sqrt{2g\Delta h}  \]

\[      = \sqrt{2 \cdot (-9.8) \cdot (-200)}  \]

\[     = 62.61 \ \mathrm{m/s}  \]

Now, we calculate the deceleration of the woman by the cushion using the same equation. The final velocity is zero while the initial velocity is 62.61 m/s. The final thickness should be (0.500-2.00) m. Thus,

\[  v^2_f – v^2_0 = 2ad  \]

\[  0 – 62.61^2 = 2a (0.5 – 2)  \]

\[  a = \frac{-62.61^2}{2 \cdot (-1.5)}  \]

\[  a = 1307 \ \mathrm{m/s}^2  \]