Empire state building physics
By bdavis
· 5,460 views
i have always been curious how it would be to drop an object from the top of the empire state building. It is obviously a long way down but exactly how fast would an object be traveling once it hit the ground? If i were to drop a golf ball for example, how fast would that travel? Well we can do this using my knowledge of one dimensional motion, a key physics concept. Acceleration due to gravity is -9.81 m/s for any object no matter the mass. Using our kinematic equations we can find out the final velocity any object will attain once it reaches the bottom of the Empire State building. The height of the empire state building is 443 meters tall. That will be our delta Y. Our acceleration is gravity which is -9.81 m/s. Our initial velocity is 0 m/s because we are dropping it from rest. We don't need the time in order to find the final velocity so we will use the equation Vf^2=Vo^2 + 2ay. That will be Vf^2=0 + 2(-9.8)(-443). Vf=93.18 m/s. That would be really fast and could seriously hurt someone. That is really cool!
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