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King Kong carries his chapstick up the 321m tall Empire State Building. At the top of the building, he drops it (oh no!). How fast will the chapstick be moving when it hits the ground?

KingChris  Dec 20, 2017
 #1
avatar+91027 
+1

We need  this formula

 

vf^2  =  vi^2  +  2 (a)(d)

 

Where  

 

vi  is the initial velocity  = 0

a  =  acceleration due to gravity  =  -9/81m/s^2

d  = the displacement  =  321 m

vf   =  the final  velocity  ......so we have....

 

vf^2  = 0  +  2 (-9.81m/s^2)(321m)

 

vf^2  =   6298.02 m^2/s^2           take the square root

 

vf  ≈  79.36 m/s

 

 

cool cool cool

CPhill  Dec 20, 2017

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