# physics

A 57.0 kg diver dives from a height of 15.0 m. She reaches a speed of 14.0 m/s just before entering the water. (a) What was the average force of air resistance (e.g., friction) acting on the diver. (b) What is the force of friction underwater if she reaches a depth of 2.5 m before stopping? Do not neglect the buoyant force of 500 N acting on the diver once underwater

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1. (a) Without air friction, she would achieve a velocity of
V = sqrt(2gH) = 17.15 m/s when entering the water. Instead, it is 14.0 m/s.
(Average force)*(distance) = K.E. loss
= (57/2)[17.15^2 - 14^2] = 2793 J
Average force = 2793/15 = 186 N

(b) [(Friction force)+(buoyancy force)]*(distance below water surface ) = (K.E. when entering water)

[Ff +500]*(2.5 m) = (57/2)(14^2)
Ff + 500 = 2234 N
Ff = 1734 N is the average friction force underwater.

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2. I don't understand why u would multiply by distance underwater my friend

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