# Estimate (a) How long it took King Kong to fall straight down from the top of the Empire State Building.....?

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"?

Relevance

h = 1/2*g*t^2

v = g*t

solve for t:

t = sqrt(2*h/g)

Thus:

v = sqrt(2*g*h)

h:=380 m; g:=9.8 m/s^2;

Results:

t = 77.55 seconds

v = 82.825 m/s

• 5 years ago

(a) Use the equation Y=V(t)+1/2(a)(t)^2

Where Y = distance king kong is falling -- 380m

V = initial velocity -- 0 m/s

(a)= 9.8m/s^2 (This number is not given, but something you should know.)

t = ?

So plug everything in and you ll get 380 = 0 +1/2(9.8)(t^2)

t= 8.8s

(b)

Now that you have found time, you want to find his velocity before landing.

Use the equation

v2=v1 +at

Where v2 is the final velocity

V1 is the initial velocity

a is 9.8 m/s^2

and t is what you have found in part A. (8.8s)

When you plug everything in, you will have:

V2 = 0 + (9.8 m/s^2)(8.8s)

V2 = 86.24m/s