# Applied Physics

An object is dropped from a height of 49 meters. The object takes 3.2s to hit the ground. The final velocity of the object when it hits the ground is -31 m/s

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asked by EJ
1. d = -(1/2) g t^2
-49 = -(1/2)(9.8)t^2
t^2 = 10
t = 3.16 seconds

v = -gt = -9.8*3.16 = -31 m/s

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posted by Damon

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