Estimate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before “landing.”

When an object is falling only under the influence of gravity, the situation is called free fall. The acceleration of a body under free fall is$9.8{\mathrm{ms}}^{-2}$.

The height of the building,$h=380\mathrm{m}$

The initial velocity of King Kong,$u=0{\mathrm{ms}}^{-1}$

Assume that the time taken to fall is $t_{\circ }$, the velocity just before landing is $v_{\circ }$, and acceleration due to gravity is $\mathrm{g}=9.8{\mathrm{ms}}^{-2}$.

From the second equation of motion, you can write the displacement of King Kong as.

$h=u\times t+\frac{1}{2}\mathrm{g}{t}_{\circ }^2$

Substituting the values in the above equation, you get.

$380=0+\frac{1}{2}\times 9.8\times t_{\circ }^2$

Solving the above equation for$t_{\circ }$, you get.

$t_{\circ }=8.81\mathrm{s}$

Thus, the time taken by King Kong to fall from the building is 8.81 s.

From the first equation of motion, you can write the velocity of King Kong at any time t as.

$v=u+\mathrm{g}t$

The velocity just before landing will be the velocity at time $t_{\circ }$.

Thus,

$\begin{array}{c}{v}_{\circ }=0+9.8\times 8.81\\ =86.34{\mathrm{ms}}^{-1}\end{array}$

Thus, the velocity of King Kong just before landing will be $86.34{\mathrm{ms}}^{-1}$.