Question

The empire state building is 1,450 feet tall. King Kong weighs 20 tons and he climbs to the very top. If he jumps off the top, what will his velocity be when he reahces the base of the building

Answer 1

Answer:velocity=$93.12m/s$

Explanation:

We shall use conservation of energy to solve this problem

we have

Potential energy of King Kong at top of building = His kinetic energy at the bottom

Potential energy of an object = $m\times g\times h$ = 20000 kg

Where m is mass of an object(20000 kg)

g is accleration due to gravity = 9.81$m/s^{2}$

h is the height above the surface of earth = 1450 feet = 441.96 meters

Applying values we get

$(P.E)_{TOP}=(20000\times 9.81\times 441.96) Joules\\\\(P.E)_{TOP}= 86712.552KiloJoules$......................(i)

Now Kinetic energy is given by

$(K.E)_{Bottom}=\frac{1}{2}mv^{2}$.......................(ii)

Equating i and ii we get

86712.552 =$\frac{1}{2}mv^{2}$

$v=\sqrt{\frac{2E}{m}}$

Applying values we get

$v=\sqrt{\frac{2\times 86712.552\times 10^{3}}{20\times 10^{3}}}\\\\v= 93.12m/s$