# applied physics

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a sky diver jumps out of a plane at a height of 5000 ft. if her parachute does not open until she reaches 1000 ft, what is her velocity at that point if air resisitance is disreguared?

• applied physics -

V^2 = Vo^2 + 2g*d
V^2 = 0 + 64*(5000-1000) = 256,000
V = 506 Ft/s

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