PLS HELP ILL MARK YOU BRAINIEST 3. 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. What was the average force of air resistance (e. g., friction)
acting on the diver?
Bonus: 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.

(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.