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3 years ago

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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, w

hat will his velocity be when he reahces the base of the building

1 answer:

lozanna [386]3 years ago

3 0

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

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

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I have made a sketch for you, and added it as attachment.

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3 years ago