Estimate (a) how long it took King Kong to fall straight down from the top of the Empire

Step 1 of 2

Part a

We are required to calculate the time required by King Kong to fall to the ground.

For vertical motion, we can consider the equation \(y=y_{0}+v_{0} t+\frac{1}{2} a t^{2}\)..

Here, \(y_{0}=0\)

\(v_{0}=0\)

So, \(y=\frac{1}{2} a t^{2}\)

Given:

\(y=380 \mathrm{~m} / \mathrm{s}^{2}\)

For vertical motion, \(a=9.80 \mathrm{~m} / \mathrm{s}^{2}\)

Substitute in \(y=\frac{1}{2} a t^{2}\)

\(\begin{array}{l}
380=\frac{1}{2} \times 9.80 \times t^{2} \\
t=\sqrt{\frac{2 \times 380}{9.80}} \mathrm{~s} \\
t=8.8 \mathrm{~s}
\end{array}\)

Therefore, the time required is approximately 8.8 s.